Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
© The McGraw−Hill
Companies, 2008
14. Spur and Helical Gears
14
711
Spur and Helical Gears
Chapter Outline
14–1
The Lewis Bending Equation 714
14–2
Surface Durability 723
14–3
AGMA Stress Equations 725
14–4
AGMA Strength Equations 727
14–5
Geometry Factors I and J (ZI and YJ) 731
14–6
The Elastic Coefficient Cp (ZE) 736
14–7
Dynamic Factor Kv 736
14–8
Overload Factor Ko 738
14–9
Surface Condition Factor Cf (ZR) 738
14–10
Size Factor Ks
14–11
Load-Distribution Factor Km (KH) 739
14–12
Hardness-Ratio Factor CH 741
14–13
Stress Cycle Life Factors YN and ZN 742
14–14
Reliability Factor KR (YZ) 743
14–15
Temperature Factor KT (Yθ) 744
14–16
Rim-Thickness Factor KB 744
14–17
Safety Factors SF and SH 745
14–18
Analysis
14–19
Design of a Gear Mesh 755
739
745
713
712
714
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
Mechanical Engineering Design
This chapter is devoted primarily to analysis and design of spur and helical gears to resist
bending failure of the teeth as well as pitting failure of tooth surfaces. Failure by bending will occur when the significant tooth stress equals or exceeds either the yield strength
or the bending endurance strength. A surface failure occurs when the significant contact
stress equals or exceeds the surface endurance strength. The first two sections present a
little of the history of the analyses from which current methodology developed.
The American Gear Manufacturers Association1 (AGMA) has for many years been
the responsible authority for the dissemination of knowledge pertaining to the design
and analysis of gearing. The methods this organization presents are in general use in the
United States when strength and wear are primary considerations. In view of this fact it
is important that the AGMA approach to the subject be presented here.
The general AGMA approach requires a great many charts and graphs—too many
for a single chapter in this book. We have omitted many of these here by choosing a
single pressure angle and by using only full-depth teeth. This simplification reduces
the complexity but does not prevent the development of a basic understanding of the
approach. Furthermore, the simplification makes possible a better development of the
fundamentals and hence should constitute an ideal introduction to the use of the general
AGMA method.2 Sections 14–1 and 14–2 are elementary and serve as an examination of
the foundations of the AGMA method. Table 14–1 is largely AGMA nomenclature.
14–1
The Lewis Bending Equation
Wilfred Lewis introduced an equation for estimating the bending stress in gear teeth in
which the tooth form entered into the formulation. The equation, announced in 1892,
still remains the basis for most gear design today.
To derive the basic Lewis equation, refer to Fig. 14–1a, which shows a cantilever
of cross-sectional dimensions F and t, having a length l and a load W t, uniformly distributed across the face width F. The section modulus I /c is Ft 2 /6, and therefore the
bending stress is
σ =
6W t l
M
=
I /c
Ft 2
(a)
Gear designers denote the components of gear-tooth forces as Wt , Wr , Wa or W t , W r ,
W a interchangeably. The latter notation leaves room for post-subscripts essential to freet
is the transmitted force of
body diagrams. For instance, for gears 2 and 3 in mesh, W23
1
500 Montgomery Street, Suite 350, Alexandria, VA 22314-1560.
2
The standards ANSI/AGMA 2001-D04 (revised AGMA 2001-C95) and ANSI/AGMA 2101-D04 (metric
edition of ANSI/AGMA 2001-D04), Fundamental Rating Factors and Calculation Methods for Involute
Spur and Helical Gear Teeth, are used in this chapter. The use of American National Standards is
completely voluntary; their existence does not in any respect preclude people, whether they have approved
the standards or not, from manufacturing, marketing, purchasing, or using products, processes, or
procedures not conforming to the standards.
The American National Standards Institute does not develop standards and will in no circumstances
give an interpretation of any American National Standard. Requests for interpretation of these standards
should be addressed to the American Gear Manufacturers Association. [Tables or other self-supporting
sections may be quoted or extracted in their entirety. Credit line should read: “Extracted from ANSI/AGMA
Standard 2001-D04 or 2101-D04 Fundamental Rating Factors and Calculation Methods for Involute Spur
and Helical Gear Teeth” with the permission of the publisher, American Gear Manufacturers Association,
500 Montgomery Street, Suite 350, Alexandria, Virginia 22314-1560.] The foregoing is adapted in part from
the ANSI foreword to these standards.
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
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713
© The McGraw−Hill
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14. Spur and Helical Gears
Spur and Helical Gears
Table 14–1
Symbols, Their Names,
and Locations∗
Symbol
Name
Where Found
b
Net width of face of narrowest member
Eq. (14–16)
Ce
Mesh alignment correction factor
Eq. (14–35)
Cf
Surface condition factor
Eq. (14–16)
CH
Hardness-ratio factor
Eq. (14–18)
Cma
Mesh alignment factor
Eq. (14–34)
Cmc
Load correction factor
Eq. (14–31)
Cmf
Face load-distribution factor
Eq. (14–30)
Cp
Elastic coefficient
Eq. (14–13)
Cpf
Pinion proportion factor
Eq. (14–32)
Cpm
Pinion proportion modifier
Eq. (14–33)
d
Operating pitch diameter of pinion
Ex. (14–1)
dP
Pitch diameter, pinion
Eq. (14–22)
dG
Pitch diameter, gear
Eq. (14–22)
E
Modulus of elasticity
Eq. (14–10)
F
Net face width of narrowest member
Eq. (14–15)
fP
Pinion surface finish
Fig. 14–13
H
Power
Fig. 14–17
Ex. 14–3
HB
Brinell hardness
HBG
Brinell hardness of gear
Sec. 14–12
HBP
Brinell hardness of pinion
Sec. 14–12
hp
Horsepower
Ex. 14–1
ht
Gear-tooth whole depth
Sec. 14–16
I
Geometry factor of pitting resistance
Eq. (14–16)
J
Geometry factor for bending strength
Eq. (14–15)
K
Contact load factor for pitting resistance
Eq. (6–65)
KB
Rim-thickness factor
Eq. (14–40)
Eq. (14–9)
Kf
Fatigue stress-concentration factor
Km
Load-distribution factor
Eq. (14–30)
Ko
Overload factor
Eq. (14–15)
KR
Reliability factor
Eq. (14–17)
Ks
Size factor
Sec. 14–10
KT
Temperature factor
Eq. (14–17)
Kv
Dynamic factor
Eq. (14–27)
m
Metric module
Eq. (14–15)
mB
Backup ratio
Eq. (14–39)
mG
Gear ratio (never less than 1)
Eq. (14–22)
mN
Load-sharing ratio
Eq. (14–21)
N
Number of stress cycles
Fig. 14–14
NG
Number of teeth on gear
Eq. (14–22)
NP
Number of teeth on pinion
Eq. (14–22)
Speed
Ex. 14–1
n
715
(Continued)
714
716
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
Mechanical Engineering Design
Table 14–1
Symbols, Their Names,
and Locations∗
(Continued)
Symbol
Name
Where Found
nP
Pinion speed
Ex. 14–4
P
Diametral pitch
Eq. (14–2)
Eq. (14–15)
Pd
Diametral pitch of pinion
pN
Normal base pitch
Eq. (14–24)
pn
Normal circular pitch
Eq. (14–24)
px
Axial pitch
Eq. (14–19)
Qv
Transmission accuracy level number
Eq. (14–29)
Eq. (14–38)
R
Reliability
Ra
Root-mean-squared roughness
Fig. 14–13
rf
Tooth fillet radius
Fig. 14–1
rG
Pitch-circle radius, gear
In standard
rP
Pitch-circle radius, pinion
In standard
rbP
Pinion base-circle radius
Eq. (14–25)
rbG
Gear base-circle radius
Eq. (14–25)
SC
Buckingham surface endurance strength
Ex. 14–3
Sc
AGMA surface endurance strength
Eq. (14–18)
St
AGMA bending strength
Eq. (14–17)
S
Bearing span
Fig. 14–10
S1
Pinion offset from center span
Fig. 14–10
SF
Safety factor—bending
Eq. (14–41)
SH
Safety factor—pitting
Eq. (14–42)
Transmitted load
Fig. 14–1
W t or W †t
YN
Stress cycle factor for bending strength
Fig. 14–14
ZN
Stress cycle factor for pitting resistance
Fig. 14–15
Exponent
Eq. (14–44)
σ
Bending stress
Eq. (14–2)
σC
Contact stress from Hertzian relationships
Eq. (14–14)
σc
Contact stress from AGMA relationships
Eq. (14–16)
σall
Allowable bending stress
Eq. (14–17)
β
σc,all
Allowable contact stress, AGMA
Eq. (14–18)
φ
Pressure angle
Eq. (14–12)
φt
Transverse pressure angle
Eq. (14–23)
ψ
Helix angle at standard pitch diameter
Ex. 14–5
∗ Because
ANSI/AGMA 2001-C95 introduced a significant amount of new nomenclature, and continued in ANSI/AGMA 2001-D04,
this summary and references are provided for use until the reader’s vocabulary has grown.
†
See preference rationale following Eq. (a), Sec. 14–1.
t
is the transmitted force of body 3 on body 2. When working
body 2 on body 3, and W32
with double- or triple-reduction speed reducers, this notation is compact and essential to
clear thinking. Since gear-force components rarely take exponents, this causes no complication. Pythagorean combinations, if necessary, can be treated with parentheses or
avoided by expressing the relations trigonometrically.
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
© The McGraw−Hill
Companies, 2008
14. Spur and Helical Gears
Spur and Helical Gears
Figure 14–1
Wr
715
717
W
Wt
l
Wt
F
rf
a
t
x
t
l
(a)
(b)
Referring now to Fig. 14–1b, we assume that the maximum stress in a gear tooth
occurs at point a. By similar triangles, you can write
l
t/2
=
x
t/2
or
x=
t2
4l
(b)
By rearranging Eq. (a),
σ =
6W t l
Wt 1 1
Wt 1
=
=
2
2
Ft
F t /6l
F t 2 /4l 46
(c)
If we now substitute the value of x from Eq. (b) in Eq. (c) and multiply the numerator
and denominator by the circular pitch p, we find
σ =
Wt p
2
F 3 xp
(d)
Letting y = 2x/3 p, we have
σ =
Wt
F py
(14–1)
This completes the development of the original Lewis equation. The factor y is called
the Lewis form factor, and it may be obtained by a graphical layout of the gear tooth or
by digital computation.
In using this equation, most engineers prefer to employ the diametral pitch in
determining the stresses. This is done by substituting P = π/ p and Y = π y in
Eq. (14–1). This gives
σ =
Wt P
FY
(14–2)
Y =
2x P
3
(14–3)
where
The use of this equation for Y means that only the bending of the tooth is considered
and that the compression due to the radial component of the force is neglected. Values
of Y obtained from this equation are tabulated in Table 14–2.
716
718
Budynas−Nisbett: Shigley’s
Mechanical Engineering
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III. Design of Mechanical
Elements
© The McGraw−Hill
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14. Spur and Helical Gears
Mechanical Engineering Design
Table 14–2
Values of the Lewis Form
Factor Y (These Values
Are for a Normal
Pressure Angle of 20°,
Full-Depth Teeth, and a
Diametral Pitch of Unity
in the Plane of Rotation)
Number of
Teeth
Y
Number of
Teeth
12
0.245
28
0.353
13
0.261
30
0.359
14
0.277
34
0.371
15
0.290
38
0.384
16
0.296
43
0.397
17
0.303
50
0.409
18
0.309
60
0.422
19
0.314
75
0.435
20
0.322
100
0.447
21
0.328
150
0.460
22
0.331
300
0.472
24
0.337
400
0.480
26
0.346
Rack
0.485
Y
The use of Eq. (14–3) also implies that the teeth do not share the load and that the
greatest force is exerted at the tip of the tooth. But we have already learned that the contact ratio should be somewhat greater than unity, say about 1.5, to achieve a quality
gearset. If, in fact, the gears are cut with sufficient accuracy, the tip-load condition is
not the worst, because another pair of teeth will be in contact when this condition
occurs. Examination of run-in teeth will show that the heaviest loads occur near the
middle of the tooth. Therefore the maximum stress probably occurs while a single pair
of teeth is carrying the full load, at a point where another pair of teeth is just on the
verge of coming into contact.
Dynamic Effects
When a pair of gears is driven at moderate or high speed and noise is generated, it is
certain that dynamic effects are present. One of the earliest efforts to account for an
increase in the load due to velocity employed a number of gears of the same size, material, and strength. Several of these gears were tested to destruction by meshing and
loading them at zero velocity. The remaining gears were tested to destruction at various
pitch-line velocities. For example, if a pair of gears failed at 500 lbf tangential load at
zero velocity and at 250 lbf at velocity V1 , then a velocity factor, designated K v , of 2
was specified for the gears at velocity V1 . Then another, identical, pair of gears running
at a pitch-line velocity V1 could be assumed to have a load equal to twice the tangential or transmitted load.
Note that the definition of dynamic factor K v has been altered. AGMA standards
ANSI/AGMA 2001-D04 and 2101-D04 contain this caution:
Dynamic factor K v has been redefined as the reciprocal of that used in previous AGMA
standards. It is now greater than 1.0. In earlier AGMA standards it was less than 1.0.
Care must be taken in referring to work done prior to this change in the standards.
Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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© The McGraw−Hill
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14. Spur and Helical Gears
Spur and Helical Gears
717
719
In the nineteenth century, Carl G. Barth first expressed the velocity factor, and in
terms of the current AGMA standards, they are represented as
Kv =
600 + V
600
(cast iron, cast profile)
(14–4a)
Kv =
1200 + V
1200
(cut or milled profile)
(14–4b)
where V is the pitch-line velocity in feet per minute. It is also quite probable, because
of the date that the tests were made, that the tests were conducted on teeth having a
cycloidal profile instead of an involute profile. Cycloidal teeth were in general use in the
nineteenth century because they were easier to cast than involute teeth. Equation (14–4a)
is called the Barth equation. The Barth equation is often modified into Eq. (14–4b), for
cut or milled teeth. Later AGMA added
√
50 + V
(hobbed or shaped profile)
Kv =
(14–5a)
50
√
78 + V
(shaved or ground profile)
Kv =
(14–5b)
78
In SI units, Eqs. (14–4a) through (14–5b) become
Kv =
3.05 + V
3.05
(cast iron, cast profile)
6.1 + V
(cut or milled profile)
6.1
√
3.56 + V
(hobbed or shaped profile)
Kv =
3.56
√
5.56 + V
(shaved or ground profile)
Kv =
5.56
Kv =
(14–6a)
(14–6b)
(14–6c)
(14–6d)
where V is in meters per second (m/s).
Introducing the velocity factor into Eq. (14–2) gives
σ =
Kv W t P
FY
(14–7)
Kv W t
FmY
(14–8)
The metric version of this equation is
σ =
where the face width F and the module m are both in millimeters (mm). Expressing
the tangential component of load W t in newtons (N) then results in stress units of
megapascals (MPa).
As a general rule, spur gears should have a face width F from 3 to 5 times the
circular pitch p.
Equations (14–7) and (14–8) are important because they form the basis for the
AGMA approach to the bending strength of gear teeth. They are in general use for
718
720
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
Mechanical Engineering Design
estimating the capacity of gear drives when life and reliability are not important considerations. The equations can be useful in obtaining a preliminary estimate of gear
sizes needed for various applications.
EXAMPLE 14–1
A stock spur gear is available having a diametral pitch of 8 teeth/in, a 1 12 -in face, 16
teeth, and a pressure angle of 20◦ with full-depth teeth. The material is AISI 1020 steel
in as-rolled condition. Use a design factor of n d = 3 to rate the horsepower output of the
gear corresponding to a speed of 1200 rev/m and moderate applications.
Solution
The term moderate applications seems to imply that the gear can be rated by using the
yield strength as a criterion of failure. From Table A–20, we find Sut = 55 kpsi and
Sy = 30 kpsi. A design factor of 3 means that the allowable bending stress is 30/3 =
10 kpsi. The pitch diameter is N/P = 16/8 = 2 in, so the pitch-line velocity is
V =
π(2)1200
πdn
=
= 628 ft/min
12
12
The velocity factor from Eq. (14–4b) is found to be
Kv =
1200 + 628
1200 + V
=
= 1.52
1200
1200
Table 14–2 gives the form factor as Y = 0.296 for 16 teeth. We now arrange and substitute in Eq. (14–7) as follows:
Wt =
1.5(0.296)10 000
FY σall
=
= 365 lbf
Kv P
1.52(8)
The horsepower that can be transmitted is
Answer
hp =
365(628)
WtV
=
= 6.95 hp
33 000
33 000
It is important to emphasize that this is a rough estimate, and that this approach
must not be used for important applications. The example is intended to help you understand some of the fundamentals that will be involved in the AGMA approach.
EXAMPLE 14–2
Solution
Estimate the horsepower rating of the gear in the previous example based on obtaining
an infinite life in bending.
The rotating-beam endurance limit is estimated from Eq. (6–8)
Se = 0.5Sut = 0.5(55) = 27.5 kpsi
To obtain the surface finish Marin factor ka we refer to Table 6–3 for machined surface,
finding a = 2.70 and b = −0.265. Then Eq. (6–19) gives the surface finish Marin
factor ka as
b
ka = aSut
= 2.70(55)−0.265 = 0.934
Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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14. Spur and Helical Gears
719
Spur and Helical Gears
721
The next step is to estimate the size factor kb . From Table 13–1, the sum of the addendum and dedendum is
l=
1.25
1 1.25
1
+
= +
= 0.281 in
P
P
8
8
The tooth thickness t in Fig. 14–1b is given in Sec. 14–1 [Eq. (b)] as t = (4lx)1/2
when x = 3Y/(2P) from Eq. (14–3). Therefore, since from Ex. 14–1 Y = 0.296 and
P = 8,
x=
3(0.296)
3Y
=
= 0.0555 in
2P
2(8)
then
t = (4lx)1/2 = [4(0.281)0.0555]1/2 = 0.250 in
We have recognized the tooth as a cantilever beam of rectangular cross section, so the
equivalent rotating-beam diameter must be obtained from Eq. (6–25):
de = 0.808(hb)1/2 = 0.808(Ft)1/2 = 0.808[1.5(0.250)]1/2 = 0.495 in
Then, Eq. (6–20) gives kb as
de −0.107
0.495 −0.107
=
= 0.948
kb =
0.30
0.30
The load factor kc from Eq. (6–26) is unity. With no information given concerning
temperature and reliability we will set kd = ke = 1.
Two effects are used to evaluate the miscellaneous-effects Marin factor k f . The
first of these is the effect of one-way bending. In general, a gear tooth is subjected
only to one-way bending. Exceptions include idler gears and gears used in reversing
mechanisms.
For one-way bending the steady and alternating stress components are σa = σm =
σ/2 where σ is the largest repeatedly applied bending stress as given in Eq. (14–7). If
a material exhibited a Goodman failure locus,
Sm
Sa
+
=1
Se
Sut
Since Sa and Sm are equal for one-way bending, we substitute Sa for Sm and solve the
preceding equation for Sa , giving
Sa =
Se Sut
Se + Sut
Now replace Sa with σ/2, and in the denominator replace Se with 0.5Sut to obtain
σ =
2Se
2Se Sut
= 1.33Se
=
0.5Sut + Sut
0.5 + 1
Now k f = σ/Se = 1.33Se /Se = 1.33. However, a Gerber fatigue locus gives mean
values of
2
Sm
Sa
+
=1
Se
Sut
720
722
Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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© The McGraw−Hill
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14. Spur and Helical Gears
Mechanical Engineering Design
Setting Sa = Sm and solving the quadratic in Sa gives
2
Sut
4Se2
Sa = −1 + 1 + 2
2Se
Sut
Setting Sa = σ/2, Sut = Se /0.5 gives
σ =
Se
−1 +
0.52
1 + 4(0.5)2 = 1.66Se
and k f = σ/Se = 1.66. Since a Gerber locus runs in and among fatigue data and
Goodman does not, we will use k f = 1.66.
The second effect to be accounted for in using the miscellaneous-effects Marin
factor k f is stress concentration, for which we will use our fundamentals from Chap. 6.
For a 20◦ full-depth tooth the radius of the root fillet is denoted r f , where
rf =
0.300
0.300
=
= 0.0375 in
P
8
From Fig. A–15–6
rf
0.0375
r
=
=
= 0.15
d
t
0.250
Since D/d = ∞, we approximate with D/d = 3, giving K t = 1.68. From Fig. 6–20,
q = 0.62. From Eq. (6–32)
K f = 1 + (0.62)(1.68 − 1) = 1.42
The miscellaneous-effects Marin factor for stress concentration can be expressed as
kf =
1
1
= 0.704
=
Kf
1.42
The final value of k f is the product of the two k f factors, that is, 1.66(0.704) = 1.17. The
Marin equation for the fully corrected endurance strength is
Se = ka kb kc kd ke k f Se
= 0.934(0.948)(1)(1)(1)1.17(27.5) = 28.5 kpsi
For a design factor of n d = 3, as used in Ex. 14–1, applied to the load or strength, the
allowable bending stress is
Se
28.5
=
= 9.5 kpsi
σall =
nd
3
The transmitted load W t is
Wt =
1.5(0.296)9 500
FY σall
=
= 347 lbf
Kv P
1.52(8)
and the power is, with V = 628 ft/min from Ex. 14–1,
hp =
347(628)
WtV
=
= 6.6 hp
33 000
33 000
Again, it should be emphasized that these results should be accepted only as preliminary estimates to alert you to the nature of bending in gear teeth.
Budynas−Nisbett: Shigley’s
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Design, Eighth Edition
III. Design of Mechanical
Elements
© The McGraw−Hill
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14. Spur and Helical Gears
Spur and Helical Gears
721
723
In Ex. 14–2 our resources (Fig. A–15–6) did not directly address stress concentration in gear teeth. A photoelastic investigation by Dolan and Broghamer reported in
1942 constitutes a primary source of information on stress concentration.3 Mitchiner
and Mabie4 interpret the results in term of fatigue stress-concentration factor K f as
L M
t
t
Kf = H +
(14–9)
r
l
where
H = 0.34 − 0.458 366 2φ
L = 0.316 − 0.458 366 2φ
M = 0.290 + 0.458 366 2φ
(b − r f )2
r=
(d/2) + b − r f
In these equations l and t are from the layout in Fig. 14–1, φ is the pressure angle, r f is
the fillet radius, b is the dedendum, and d is the pitch diameter. It is left as an exercise
for the reader to compare K f from Eq. (14–9) with the results of using the approximation of Fig. A–15–6 in Ex. 14–2.
14–2
Surface Durability
In this section we are interested in the failure of the surfaces of gear teeth, which is
generally called wear. Pitting, as explained in Sec. 6–16, is a surface fatigue failure due to
many repetitions of high contact stresses. Other surface failures are scoring, which is a lubrication failure, and abrasion, which is wear due to the presence of foreign material.
To obtain an expression for the surface-contact stress, we shall employ the Hertz
theory. In Eq. (3–74) it was shown that the contact stress between two cylinders may be
computed from the equation
pmax =
where
2F
πbl
(a)
pmax = largest surface pressure
F = force pressing the two cylinders together
l = length of cylinders
and half-width b is obtained from Eq. (3–73):
1/2
2F 1 − ν12 E 1 + 1 − ν22 E 2
b=
πl
(1/d1 ) + (1/d2 )
(14–10)
where ν1 , ν2 , E 1 , and E 2 are the elastic constants and d1 and d2 are the diameters,
respectively, of the two contacting cylinders.
To adapt these relations to the notation used in gearing, we replace F by W t /cos φ,
d by 2r, and l by the face width F. With these changes, we can substitute the value of b
3
T. J. Dolan and E. I. Broghamer, A Photoelastic Study of the Stresses in Gear Tooth Fillets, Bulletin 335,
Univ. Ill. Exp. Sta., March 1942, See also W. D. Pilkey, Peterson’s Stress Concentration Factors, 2nd ed.,
John Wiley & Sons, New York, 1997, pp. 383–385, 412–415.
4
R. G. Mitchiner and H. H. Mabie, “Determination of the Lewis Form Factor and the AGMA Geometry
Factor J of External Spur Gear Teeth,” J. Mech. Des., Vol. 104, No. 1, Jan. 1982, pp. 148–158.
722
724
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14. Spur and Helical Gears
Mechanical Engineering Design
as given by Eq. (14–10) in Eq. (a). Replacing pmax by σC , the surface compressive
stress (Hertzian stress) is found from the equation
σC2 =
Wt
(1/r1 ) + (1/r2 )
2
π F cos φ 1 − ν1 E 1 + 1 − ν22 E 2
(14–11)
where r1 and r2 are the instantaneous values of the radii of curvature on the pinion- and
gear-tooth profiles, respectively, at the point of contact. By accounting for load sharing
in the value of W t used, Eq. (14–11) can be solved for the Hertzian stress for any or
all points from the beginning to the end of tooth contact. Of course, pure rolling exists
only at the pitch point. Elsewhere the motion is a mixture of rolling and sliding.
Equation (14–11) does not account for any sliding action in the evaluation of stress. We
note that AGMA uses μ for Poisson’s ratio instead of ν as is used here.
We have already noted that the first evidence of wear occurs near the pitch line. The
radii of curvature of the tooth profiles at the pitch point are
r1 =
d P sin φ
2
r2 =
dG sin φ
2
(14–12)
where φ is the pressure angle and d P and dG are the pitch diameters of the pinion and
gear, respectively.
Note, in Eq. (14–11), that the denominator of the second group of terms contains
four elastic constants, two for the pinion and two for the gear. As a simple means of combining and tabulating the results for various combinations of pinion and gear materials,
AGMA defines an elastic coefficient C p by the equation
⎡
⎤1/2
⎢
⎥
1
⎢
⎥
Cp = ⎢ ⎥
⎣
1 − νG2 ⎦
1 − ν 2P
π
+
EP
EG
(14–13)
With this simplification, and the addition of a velocity factor K v , Eq. (14–11) can be
written as
1 1/2
1
Kv W t
+
σC = −C p
(14–14)
F cos φ r1 r2
where the sign is negative because σC is a compressive stress.
EXAMPLE 14–3
The pinion of Examples 14–1 and 14–2 is to be mated with a 50-tooth gear manufactured of ASTM No. 50 cast iron. Using the tangential load of 382 lbf, estimate the factor
of safety of the drive based on the possibility of a surface fatigue failure.
Solution
From Table A–5 we find the elastic constants to be E P = 30 Mpsi, ν P = 0.292, E G =
14.5 Mpsi, νG = 0.211. We substitute these in Eq. (14–13) to get the elastic coefficient as
⎫1/2
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎨
1
= 1817
Cp =
⎪
1 − (0.211)2 ⎪
1 − (0.292)2
⎪
⎪
⎪
⎪
+
π
⎭
⎩
30(106 )
14.5(106 )
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14. Spur and Helical Gears
Spur and Helical Gears
723
725
From Example 14–1, the pinion pitch diameter is d P = 2 in. The value for the gear is
dG = 50/8 = 6.25 in. Then Eq. (14–12) is used to obtain the radii of curvature at the
pitch points. Thus
r1 =
2 sin 20◦
= 0.342 in
2
r2 =
6.25 sin 20◦
= 1.069 in
2
The face width is given as F = 1.5 in. Use K v = 1.52 from Example 14–1. Substituting
all these values in Eq. (14–14) with φ = 20◦ gives the contact stress as
1/2
1
1.52(380)
1
+
= −72 400 psi
σC = −1817
1.5 cos 20◦ 0.342 1.069
The surface endurance strength of cast iron can be estimated from
SC = 0.32HB kpsi
for 10 cycles, where SC is in kpsi. Table A–24 gives HB = 262 for ASTM No. 50 cast
iron. Therefore SC = 0.32(262) = 83.8 kpsi. Contact stress is not linear with transmitted load [see Eq. (14–14)]. If the factor of safety is defined as the loss-of-function load
divided by the imposed load, then the ratio of loads is the ratio of stresses squared. In
other words,
SC2
83.8 2
loss-of-function load
= 2 =
= 1.34
n=
imposed load
72.4
σC
8
One is free to define factor of safety as SC /σC . Awkwardness comes when one compares the factor of safety in bending fatigue with the factor of safety in surface fatigue
for a particular gear. Suppose the factor of safety of this gear in bending fatigue is 1.20
and the factor of safety in surface fatigue is 1.34 as above. The threat, since 1.34 is
greater than 1.20, is in bending fatigue since both numbers √
are based on load ratios. If
the factor of safety in surface fatigue is based on SC /σC = 1.34 = 1.16, then 1.20 is
greater than 1.16, but the threat is not from surface fatigue. The surface fatigue factor
of safety can be defined either way. One way has the burden of requiring a squared
number before numbers that instinctively seem comparable can be compared.
In addition to the dynamic factor K v already introduced, there are transmitted load
excursions, nonuniform distribution of the transmitted load over the tooth contact, and the
influence of rim thickness on bending stress. Tabulated strength values can be means,
ASTM minimums, or of unknown heritage. In surface fatigue there are no endurance limits. Endurance strengths have to be qualified as to corresponding cycle count, and the slope
of the S-N curve needs to be known. In bending fatigue there is a definite change in slope
of the S-N curve near 106 cycles, but some evidence indicates that an endurance limit does
not exist. Gearing experience leads to cycle counts of 1011 or more. Evidence of diminishing endurance strengths in bending have been included in AGMA methodology.
14–3
AGMA Stress Equations
Two fundamental stress equations are used in the AGMA methodology, one for bending stress and another for pitting resistance (contact stress). In AGMA terminology,
these are called stress numbers, as contrasted with actual applied stresses, and are
724
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Mechanical Engineering Design
designated by a lowercase letter s instead of the Greek lower case σ we have used in
this book (and shall continue to use). The fundamental equations are
⎧
Pd K m K B
⎪
t
⎪
(U.S. customary units)
⎪
⎨ W Ko Kv Ks F
J
σ =
(14–15)
1 KH KB
⎪
t
⎪
⎪
K
K
K
(SI
units)
W
o
v
s
⎩
bm t
YJ
where for U.S. customary units (SI units),
W t is the tangential transmitted load, lbf (N)
K o is the overload factor
K v is the dynamic factor
K s is the size factor
Pd is the transverse diameteral pitch
F (b) is the face width of the narrower member, in (mm)
K m (KH) is the load-distribution factor
K B is the rim-thickness factor
J (Y J ) is the geometry factor for bending strength (which includes root fillet
stress-concentration factor K f )
(m t ) is the transverse metric module
Before you try to digest the meaning of all these terms in Eq. (14–15), view them as
advice concerning items the designer should consider whether he or she follows the
voluntary standard or not. These items include issues such as
• Transmitted load magnitude
• Overload
•
•
•
•
•
Dynamic augmentation of transmitted load
Size
Geometry: pitch and face width
Distribution of load across the teeth
Rim support of the tooth
• Lewis form factor and root fillet stress concentration
The fundamental equation for pitting resistance (contact stress) is
⎧
Km C f
⎪
⎪
t
⎪
(U.S. customary units)
⎨ Cp W Ko Kv Ks d F I
P
σc =
⎪
ZR
K
⎪
⎪ Z E W t Ko Kv Ks H
(SI units)
⎩
dw1 b Z I
(14–16)
where W t, Ko, K v , Ks, Km, F, and b are the same terms as defined for Eq. (14–15). For
U.S. customary units (SI units), the additional terms are
√
√
C p (Z E ) is an elastic coefficient, lbf/in2 ( N/mm2 )
C f (Z R ) is the surface condition factor
d P (dw1 ) is the pitch diameter of the pinion, in (mm)
I (Z I ) is the geometry factor for pitting resistance
The evaluation of all these factors is explained in the sections that follow. The development of Eq. (14–16) is clarified in the second part of Sec. 14–5.
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14. Spur and Helical Gears
Spur and Helical Gears
14–4
725
727
AGMA Strength Equations
Instead of using the term strength, AGMA uses data termed allowable stress numbers
and designates these by the symbols sat and sac . It will be less confusing here if we continue the practice in this book of using the uppercase letter S to designate strength and
the lowercase Greek letters σ and τ for stress. To make it perfectly clear we shall use the
term gear strength as a replacement for the phrase allowable stress numbers as used by
AGMA.
Following this convention, values for gear bending strength, designated here as St ,
are to be found in Figs. 14–2, 14–3, and 14–4, and in Tables 14–3 and 14–4. Since gear
strengths are not identified with other strengths such as Sut , Se , or Sy as used elsewhere
in this book, their use should be restricted to gear problems.
In this approach the strengths are modified by various factors that produce limiting
values of the bending stress and the contact stress.
Figure 14–2
Metallurgical and quality
control procedure required
Allowable bending stress number, St kpsi
Allowable bending stress
number for through-hardened
steels. The SI equations
are St = 0.533HB +
88.3 MPa, grade 1, and
St = 0.703HB + 113 MPa,
grade 2. (Source:
ANSI/AGMA
2001-D04 and 2101-D04.)
50
Grade 2
St = 102 HB + 16 400 psi
40
30
Grade 1
St = 77.3 HB + 12 800 psi
20
10
150
200
250
300
350
400
450
Brinell hardness, HB
80
Figure 14–3
Metallurgical and quality control procedures required
Allowable bending stress number, St kpsi
Allowable bending stress
number for nitrided throughhardened steel gears (i.e.,
AISI 4140, 4340), St . The
SI equations are St =
0.568H B + 83.8 MPa,
grade 1, and St =
0.749H B + 110 MPa,
grade 2. (Source:
ANSI/AGMA
2001-D04 and 2101-D04.)
70
60
Grade 2
St = 108.6HB + 15 890 psi
50
40
Grade 1
St = 82.3HB + 12 150 psi
30
20
250
275
300
Core hardness, HB
325
350
726
728
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14. Spur and Helical Gears
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Figure 14–4
70
Allowable bending stress numbers, St kpsi
Metallurgical and quality control procedures required
Allowable bending stress
numbers for nitriding steel
gears St . The SI equations
are St = 0.594HB + 87.76
MPa Nitralloy grade 1
St = 0.784HB + 114.81
MPa Nitralloy grade 2
St = 0.7255HB + 63.89
MPa 2.5% chrome, grade 1
St = 0.7255HB + 153.63
MPa 2.5% chrome, grade 2
St = 0.7255HB + 201.91
MPa 2.5% chrome, grade 3
(Source: ANSI/AGMA 2001D04,2101-D04.)
Grade 3 − 2.5% Chrome
St = 105.2HB + 29 280 psi
60
Grade 2 − 2.5% Chrome
St = 105.2HB + 22 280 psi
Grade 2 − Nitralloy
St = 113.8HB + 16 650 psi
50
Grade 1 − 2.5% Chrome
St = 105.2HB + 9280 psi
40
Grade 1 − Nitralloy
St = 86.2HB + 12 730 psi
30
250
275
300
325
350
Core hardness, HB
Table 14–3
Repeatedly Applied Bending Strength St at 107 Cycles and 0.99 Reliability for Steel Gears
Source: ANSI/AGMA 2001-D04.
Material
Designation
Heat
Treatment
Steel3
Through-hardened
Flame4 or induction
hardened4 with type
A pattern5
Flame4 or induction
hardened4 with type
B pattern5
Nitralloy 135M,
Nitralloy N, and
2.5% chrome
Minimum
Surface
Hardness1
Allowable Bending Stress Number St,2
psi
Grade 1
Grade 2
Grade 3
See Fig. 14–2
See Fig. 14–2
See Fig. 14–2
—
See Table 8*
45 000
55 000
—
See Table 8*
22 000
22 000
—
Carburized and
hardened
See Table 9*
55 000
65 000 or
70 0006
75 000
Nitrided4,7 (throughhardened steels)
83.5 HR15N
See Fig. 14–3
See Fig. 14–3
—
Nitrided4,7
87.5 HR15N
See Fig. 14–4
See Fig. 14–4
See Fig. 14–4
(no aluminum)
Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–7.
1
Hardness to be equivalent to that at the root diameter in the center of the tooth space and face width.
2
See tables 7 through 10 for major metallurgical factors for each stress grade of steel gears.
3
The steel selected must be compatible with the heat treatment process selected and hardness required.
4
The allowable stress numbers indicated may be used with the case depths prescribed in 16.1.
5
See figure 12 for type A and type B hardness patterns.
6
If bainite and microcracks are limited to grade 3 levels, 70,000 psi may be used.
7
The overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design. [7]
*Tables 8 and 9 of ANSI/AGMA 2001-D04 are comprehensive tabulations of the major metallurgical factors affecting St and Sc of flame-hardened and induction-hardened (Table 8)
and carburized and hardened (Table 9) steel gears.
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14. Spur and Helical Gears
729
Spur and Helical Gears
Table 14–4
Repeatedly Applied Bending Strength St for Iron and Bronze Gears at 107 Cycles and 0.99 Reliability
Source: ANSI/AGMA 2001-D04.
Material
ASTM A48 gray
cast iron
ASTM A536 ductile
(nodular) Iron
Typical Minimum
Surface Hardness2
Allowable Bending
Stress Number, St,3
psi
As cast
—
5000
Class 30
As cast
174 HB
8500
Class 40
As cast
201 HB
13 000
Material
Designation1
Heat
Treatment
Class 20
Grade 60–40–18
Annealed
140 HB
22 000–33 000
Grade 80–55–06
Quenched and
tempered
179 HB
22 000–33 000
Grade 100–70–03
Quenched and
tempered
229 HB
27 000–40 000
Grade 120–90–02
Quenched and
tempered
269 HB
31 000–44 000
Minimum tensile strength
40 000 psi
5700
Minimum tensile strength
23 600
Bronze
Sand cast
ASTM B–148
Heat treated
Alloy 954
90 000 psi
Notes:
1
See ANSI/AGMA 2004-B89, Gear Materials and Heat Treatment Manual.
2
Measured hardness to be equivalent to that which would be measured at the root diameter in the center of the tooth space and face width.
3
The lower values should be used for general design purposes. The upper values may be used when:
High quality material is used.
Section size and design allow maximum response to heat treatment.
Proper quality control is effected by adequate inspection.
Operating experience justifies their use.
The equation for the allowable bending stress is
σall
⎧
St
YN
⎪
⎪
⎨
SF K T K R
=
⎪ St Y N
⎪
⎩
S F Yθ Y Z
(U.S. customary units)
(14–17)
(SI units)
where for U.S. customary units (SI units),
St is the allowable bending stress, lbf/in2 (N/mm2)
Y N is the stress cycle factor for bending stress
K T (Yθ ) are the temperature factors
K R (Y Z ) are the reliability factors
S F is the AGMA factor of safety, a stress ratio
728
730
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14. Spur and Helical Gears
Mechanical Engineering Design
The equation for the allowable contact stress σc,all is
⎧
Sc Z N C H
⎪
⎪
(U.S. customary units)
⎨
SH K T K R
σc,all =
(14–18)
S Z Z
⎪
⎪
⎩ c N W
(SI units)
S H Yθ Y Z
where the upper equation is in U.S. customary units and the lower equation is in SI units,
Also,
Sc is the allowable contact stress, lbf/in2 (N/mm2)
Z N is the stress cycle life factor
C H (Z W ) are the hardness ratio factors for pitting resistance
K T (Yθ ) are the temperature factors
K R (Y Z ) are the reliability factors
S H is the AGMA factor of safety, a stress ratio
The values for the allowable contact stress, designated here as Sc , are to be found in
Fig. 14–5 and Tables 14–5, 14–6, and 14–7.
AGMA allowable stress numbers (strengths) for bending and contact stress are for
• Unidirectional loading
• 10 million stress cycles
Contact-fatigue strength Sc
at 107 cycles and 0.99
reliability for throughhardened steel gears. The SI
equations are
Sc = 2.22HB + 200 MPa,
grade 1, and
Sc = 2.41HB + 237 MPa,
grade 2. (Source:
ANSI/AGMA 2001-D04 and
2101-D04.)
Allowable contact stress number, Sc
Figure 14–5
1000 lb ⁄ in2
• 99 percent reliability
Metallurgical and quality control procedures required
175
Grade 2
Sc = 349 HB + 34 300 psi
150
125
Grade 1
Sc = 322 HB + 29 100 psi
100
75
150
200
250
300
350
400
450
Brinell hardness, HB
Table 14–5
Nominal Temperature
Used in Nitriding and
Hardnesses Obtained
Source: Darle W. Dudley,
Handbook of Practical
Gear Design, rev. ed.,
McGraw-Hill,
New York, 1984.
Temperature
before nitriding, °F
Nitriding,
°F
Nitralloy 135*
1150
Nitralloy 135M
1150
Nitralloy N
AISI 4340
Steel
Hardness,
Rockwell C Scale
Case
Core
975
62–65
30–35
975
62–65
32–36
1000
975
62–65
40–44
1100
975
48–53
27–35
AISI 4140
1100
975
49–54
27–35
31 Cr Mo V 9
1100
975
58–62
27–33
∗ Nitralloy
is a trademark of the Nitralloy Corp., New York.
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14. Spur and Helical Gears
731
Spur and Helical Gears
Table 14–6
Repeatedly Applied Contact Strength Sc at 107 Cycles and 0.99 Reliability for Steel Gears
Source: ANSI/AGMA 2001-D04.
Material
Designation
3
Steel
Minimum
Surface
Hardness1
Heat
Treatment
4
Through hardened
5
See Fig. 14–5
Allowable Contact Stress Number,2 Sc, psi
Grade 1
Grade 2
Grade 3
See Fig. 14–5
See Fig. 14–5
—
—
Flame or induction
hardened5
50 HRC
170 000
190 000
54 HRC
175 000
195 000
—
Carburized and
hardened5
See Table 9∗
180 000
225 000
275 000
Nitrided5 (through
hardened steels)
83.5 HR15N
150 000
163 000
175 000
84.5 HR15N
155 000
168 000
180 000
2.5% chrome
(no aluminum)
Nitrided5
87.5 HR15N
155 000
172 000
189 000
Nitralloy 135M
Nitrided5
90.0 HR15N
170 000
183 000
195 000
Nitralloy N
5
Nitrided
90.0 HR15N
172 000
188 000
205 000
2.5% chrome
Nitrided5
90.0 HR15N
176 000
196 000
216 000
(no aluminum)
Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–5.
1
Hardness to be equivalent to that at the start of active profile in the center of the face width.
2
See Tables 7 through 10 for major metallurgical factors for each stress grade of steel gears.
3
The steel selected must be compatible with the heat treatment process selected and hardness required.
4
These materials must be annealed or normalized as a minimum.
5
The allowable stress numbers indicated may be used with the case depths prescribed in 16.1.
*Table 9 of ANSI/AGMA 2001-D04 is a comprehensive tabulation of the major metallurgical factors affecting St and Sc of carburized and hardened steel gears.
The factors in this section, too, will be evaluated in subsequent sections.
When two-way (reversed) loading occurs, as with idler gears, AGMA recommends
using 70 percent of St values. This is equivalent to 1/0.70 = 1.43 as a value of ke in
Ex. 14–2. The recommendation falls between the value of ke = 1.33 for a Goodman
failure locus and ke = 1.66 for a Gerber failure locus.
14–5
Geometry Factors I and J (ZI and YJ)
We have seen how the factor Y is used in the Lewis equation to introduce the effect of
tooth form into the stress equation. The AGMA factors5 I and J are intended to accomplish the same purpose in a more involved manner.
The determination of I and J depends upon the face-contact ratio m F . This is
defined as
mF =
F
px
(14–19)
where px is the axial pitch and F is the face width. For spur gears, m F = 0.
5
A useful reference is AGMA 908-B89, Geometry Factors for Determining Pitting Resistance and Bending
Strength of Spur, Helical and Herringbone Gear Teeth.
730
732
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14. Spur and Helical Gears
Mechanical Engineering Design
Table 14–7
Repeatedly Applied Contact Strength Sc 107 Cycles and 0.99 Reliability for Iron and Bronze Gears
Source: ANSI/AGMA 2001-D04.
Typical Minimum
Surface Hardness2
Allowable Contact
Stress Number,3 Sc,
psi
As cast
As cast
As cast
—
174 HB
201 HB
50 000–60 000
65 000–75 000
75 000–85 000
Grade 60–40–18
Grade 80–55–06
Annealed
Quenched and
tempered
140 HB
179 HB
77 000–92 000
77 000–92 000
Grade 100–70–03
Quenched and
tempered
229 HB
92 000–112 000
Grade 120–90–02
Quenched and
tempered
269 HB
103 000–126 000
—
Sand cast
Minimum tensile
strength 40 000 psi
30 000
ASTM B-148
Heat treated
Minimum tensile
65 000
Material
Designation1
Heat
Treatment
ASTM A48 gray
cast iron
Class 20
Class 30
Class 40
ASTM A536 ductile
(nodular) iron
Material
Bronze
Alloy 954
strength 90 000 psi
Notes:
1
See ANSI/AGMA 2004-B89, Gear Materials and Heat Treatment Manual.
2
Hardness to be equivalent to that at the start of active profile in the center of the face width.
3
The lower values should be used for general design purposes. The upper values may be used when:
High-quality material is used.
Section size and design allow maximum response to heat treatment.
Proper quality control is effected by adequate inspection.
Operating experience justifies their use.
Low-contact-ratio (LCR) helical gears having a small helix angle or a thin face
width, or both, have face-contact ratios less than unity (m F ≤ 1), and will not be considered here. Such gears have a noise level not too different from that for spur gears.
Consequently we shall consider here only spur gears with m F = 0 and conventional
helical gears with m F > 1.
Bending-Strength Geometry Factor J (YJ)
The AGMA factor J employs a modified value of the Lewis form factor, also denoted
by Y; a fatigue stress-concentration factor K f ; and a tooth load-sharing ratio m N . The
resulting equation for J for spur and helical gears is
J=
Y
Kf mN
(14–20)
It is important to note that the form factor Y in Eq. (14–20) is not the Lewis factor at
all. The value of Y here is obtained from calculations within AGMA 908-B89, and is
often based on the highest point of single-tooth contact.
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14. Spur and Helical Gears
733
Spur and Helical Gears
The factor K f in Eq. (14–20) is called a stress correction factor by AGMA. It is
based on a formula deduced from a photoelastic investigation of stress concentration in
gear teeth over 50 years ago.
The load-sharing ratio m N is equal to the face width divided by the minimum total
length of the lines of contact. This factor depends on the transverse contact ratio m p ,
the face-contact ratio m F , the effects of any profile modifications, and the tooth deflection. For spur gears, m N = 1.0. For helical gears having a face-contact ratio m F > 2.0,
a conservative approximation is given by the equation
mN =
pN
0.95Z
(14–21)
where p N is the normal base pitch and Z is the length of the line of action in the transverse plane (distance L ab in Fig. 13–15).
Use Fig. 14–6 to obtain the geometry factor J for spur gears having a 20◦ pressure
angle and full-depth teeth. Use Figs. 14–7 and 14–8 for helical gears having a 20◦ normal
pressure angle and face-contact ratios of m F = 2 or greater. For other gears, consult
the AGMA standard.
Gear addendum 1.000
0.55
0.50
1000
170
85
50
35
25
17
20°
Geometry factor J
0.35 rT
0.45
Generating rack 1 pitch
Load applied at highest point
of single-tooth contact
0.60
2.400
Whole depth
0.60
Addendum
1.000
Pinion addendum 1.000
Number of teeth
in mating gear
0.40
0.55
0.50
0.45
0.40
0.35
0.35
0.30
0.30
Load applied at tip of tooth
0.25
0.25
0.20
12
0.20
15
17
20
24
30
35
40 45 50
60
80
125
275
∞
Number of teeth for which geometry factor is desired
Figure 14–6
Spur-gear geometry factors J. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA
908-B89. The graph is convenient for design purposes.
732
Budynas−Nisbett: Shigley’s
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734
III. Design of Mechanical
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14. Spur and Helical Gears
Mechanical Engineering Design
Add.
1.0
Pnd
2.355
Pnd
Tooth height
Generating rack
20°
rT = 0.4276
Pnd
(a)
mN =
pN
0.95Z
Value for Z is for an element of indicated
numbers of teeth and a 75-tooth mate
Normal tooth thickness of pinion and gear
tooth each reduced 0.024 in to provide 0.048 in
total backlash for one normal diametral pitch
0.70
500
150
60
0.50
30
Number of teeth
Geometry factor J '
0.60
Factors are for
teeth cut with
a full fillet hob
20
0.40
0.30
0°
5°
10°
15°
20°
25°
30°
35°
Helix angle ␺
(b)
Figure 14–7
Helical-gear geometry factors J . Source: The graph is from AGMA 218.01, which is consistent with tabular data
from the current AGMA 908-B89. The graph is convenient for design purposes.
Surface-Strength Geometry Factor I (ZI)
The factor I is also called the pitting-resistance geometry factor by AGMA. We will
develop an expression for I by noting that the sum of the reciprocals of Eq. (14–14),
from Eq. (14–12), can be expressed as
1
2
1
1
1
+
=
+
(a)
r1 r2
sin φt d P
dG
where we have replaced φ by φt , the transverse pressure angle, so that the relation will
apply to helical gears too. Now define speed ratio m G as
mG =
NG
dG
=
NP
dP
(14–22)
III. Design of Mechanical
Elements
733
© The McGraw−Hill
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14. Spur and Helical Gears
735
Spur and Helical Gears
Figure 14–8
1.05
500
150
75
50
1.00
Modifying factor
J -factor multipliers for use
with Fig. 14–7 to find J.
Source: The graph is from
AGMA 218.01, which is
consistent with tabular data
from the current AGMA
908-B89. The graph is
convenient for design
purposes.
The modifying factor can be applied to the
J factor when other than 75 teeth are used
in the mating element
30
20
0.95
Number of teeth in mating element
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
0.90
0.85
0°
5°
10°
15°
20°
25°
30°
35°
Helix angle ␺
Equation (a) can now be written
mG + 1
1
2
1
+
=
r1 r2
d P sin φt m G
(b)
Now substitute Eq. (b) for the sum of the reciprocals in Eq. (14–14). The result is found
to be
⎤1/2
⎡
⎥
⎢ KV W t
1
⎥
σc = −σC = C p ⎢
⎣ d P F cos φt sin φt m G ⎦
2
mG + 1
(c)
The geometry factor I for external spur and helical gears is the denominator of the second term in the brackets in Eq. (c). By adding the load-sharing ratio m N , we obtain a
factor valid for both spur and helical gears. The equation is then written as
⎧
cos φt sin φt m G
⎪
⎪
external gears
⎨
2m N
mG + 1
I =
(14–23)
cos φt sin φt m G
⎪
⎪
⎩
internal gears
2m N
mG − 1
where m N = 1 for spur gears. In solving Eq. (14–21) for m N , note that
p N = pn cos φn
(14–24)
where pn is the normal circular pitch. The quantity Z, for use in Eq. (14–21), can be
obtained from the equation
1/2
2 1/2
+ (r G + a)2 − rbG
− (r P + r G ) sin φt (14–25)
Z = (r P + a)2 − rb2P
where r P and r G are the pitch radii and rb P and rbG the base-circle radii of the pinion
and gear, respectively.6 Recall from Eq. (13–6), the radius of the base circle is
rb = r cos φt
(14–26)
6
For a development, see Joseph E. Shigley and John J. Uicker Jr., Theory of Machines and Mechanisms,
McGraw-Hill, New York, 1980, p. 262.
734
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III. Design of Mechanical
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14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
Mechanical Engineering Design
Certain precautions must be taken in using Eq. (14–25). The tooth profiles are not conjugate below the base circle, and consequently, if either one or the other of the first two
terms in brackets is larger than the third term, then it should be replaced by the third
term. In addition, the effective outside radius is sometimes less than r + a, owing to
removal of burrs or rounding of the tips of the teeth. When this is the case, always use
the effective outside radius instead of r + a.
14–6
The Elastic Coefficient Cp (ZE)
Values of Cp may be computed directly from Eq. (14–13) or obtained from Table 14–8.
14–7
Dynamic Factor Kv
As noted earlier, dynamic factors are used to account for inaccuracies in the manufacture and meshing of gear teeth in action. Transmission error is defined as the departure
from uniform angular velocity of the gear pair. Some of the effects that produce transmission error are:
• Inaccuracies produced in the generation of the tooth profile; these include errors in
tooth spacing, profile lead, and runout
• Vibration of the tooth during meshing due to the tooth stiffness
• Magnitude of the pitch-line velocity
•
•
•
•
Dynamic unbalance of the rotating members
Wear and permanent deformation of contacting portions of the teeth
Gearshaft misalignment and the linear and angular deflection of the shaft
Tooth friction
In an attempt to account for these effects, AGMA has defined a set of quality numbers.7 These numbers define the tolerances for gears of various sizes manufactured to a
specified accuracy. Quality numbers 3 to 7 will include most commercial-quality gears.
Quality numbers 8 to 12 are of precision quality. The AGMA transmission accuracylevel number Q v could be taken as the same as the quality number. The following equations for the dynamic factor are based on these Q v numbers:
⎧
√ B
⎪
⎪
⎪
V
A
+
⎪
⎪
V in ft/min
⎪
⎨
A
Kv = (14–27)
B
√
⎪
⎪
⎪
A
+
200V
⎪
⎪
V in m/s
⎪
⎩
A
where
A = 50 + 56(1 − B)
(14–28)
B = 0.25(12 − Q v )2/3
and the maximum velocity, representing the end point of the Q v curve, is given by
⎧
⎪
ft/min
⎨ [A + (Q v − 3)]2
(Vt )max =
(14–29)
2
⎪
⎩ [A + (Q v − 3)]
m/s
200
7
AGMA 2000-A88. ANSI/AGMA 2001-D04, adopted in 2004, replaced Q v with Av and incorporated
ANSI/AGMA 2015-1-A01. Av ranges from 6 to 12, with lower numbers representing greater accuracy. The
Q v approach was maintained as an alternate approach, and resulting K v values are comparable.
2160
(179)
2100
(174)
1950
(162)
1900
24 × 106
(1.7 × 105)
22 × 106
(1.5 × 105)
17.5 × 106
(1.2 × 105)
16 × 106
Nodular iron
Cast iron
Aluminum bronze
Tin bronze
(154)
1850
1900
(158)
2020
(168)
2070
(172)
2090
(174)
2180
(181)
∗
(152)
1830
1880
(156)
2000
(166)
2050
(170)
2070
(172)
2160
(179)
Nodular
Iron
24 ⴛ 106
(1.7 ⴛ 105)
(149)
1800
1850
(154)
1960
(163)
2000
(166)
2020
(168)
2100
(174)
Cast
Iron
22 ⴛ 106
(1.5 ⴛ 105)
Gear Material and Modulus
of Elasticity EG, lbf/in2 (MPa)*
(141)
1700
1750
(145)
1850
(154)
1880
(156)
1900
(158)
1950
(162)
Aluminum
Bronze
17.5 ⴛ 106
(1.2 ⴛ 105)
(137)
1650
1700
(141)
1800
(149)
1830
(152)
1850
(154)
1900
(158)
Tin
Bronze
16 ⴛ 106
(1.1 ⴛ 105)
14. Spur and Helical Gears
Poisson’s ratio ⫽ 0.30.
When more exact values for modulus of elasticity are obtained from roller contact tests, they may be used.
(158)
2180
(181)
25 × 106
(1.7 × 105)
Malleable iron
(1.1 × 105)
2300
(191)
30 × 106
(2 × 105)
Steel
Pinion
Material
Malleable
Iron
25 ⴛ 106
(1.7 ⴛ 105)
Source: AGMA 218.01
Steel
30 ⴛ 106
(2 ⴛ 105)
√
psi ( MPa)
III. Design of Mechanical
Elements
Pinion Modulus of
Elasticity Ep
psi (MPa)*
Elastic Coefficient Cp (ZE),
Table 14–8
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
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Companies, 2008
735
737
736
738
Budynas−Nisbett: Shigley’s
Mechanical Engineering
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III. Design of Mechanical
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© The McGraw−Hill
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14. Spur and Helical Gears
Mechanical Engineering Design
Figure 14–9
Qv = 5
1.8
Qv = 6
1.7
Qv = 7
1.6
Dynamic factor, Kv
Dynamic factor Kv. The
equations to these curves are
given by Eq. (14–27) and the
end points by Eq. (14–29).
(ANSI/AGMA 2001-D04,
Annex A)
Qv = 8
1.5
Qv = 9
1.4
Qv = 10
1.3
1.2
Qv = 11
1.1
"Very Accurate Gearing"
1.0
0
2000
4000
6000
8000
10 000
Pitch line velocity, Vt , ft ⁄ min
Figure 14–9 is a graph of K v , the dynamic factor, as a function of pitch-line speed for
graphical estimates of K v .
14–8
Overload Factor Ko
The overload factor K o is intended to make allowance for all externally applied loads in
excess of the nominal tangential load W t in a particular application (see Figs. 14–17 and
14–18). Examples include variations in torque from the mean value due to firing of cylinders in an internal combustion engine or reaction to torque variations in a piston pump
drive. There are other similar factors such as application factor or service factor. These
factors are established after considerable field experience in a particular application.8
14–9
Surface Condition Factor Cf (ZR)
The surface condition factor C f or Z R is used only in the pitting resistance equation,
Eq. (14–16). It depends on
• Surface finish as affected by, but not limited to, cutting, shaving, lapping, grinding,
shotpeening
• Residual stress
• Plastic effects (work hardening)
Standard surface conditions for gear teeth have not yet been established. When a detrimental surface finish effect is known to exist, AGMA specifies a value of C f greater
than unity.
8
An extensive list of service factors appears in Howard B. Schwerdlin, “Couplings,” Chap. 16 in Joseph E.
Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design,
3rd ed., McGraw-Hill, New York, 2004.
Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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14. Spur and Helical Gears
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Spur and Helical Gears
14–10
737
739
Size Factor Ks
The size factor reflects nonuniformity of material properties due to size. It depends
upon
• Tooth size
• Diameter of part
• Ratio of tooth size to diameter of part
•
•
•
•
Face width
Area of stress pattern
Ratio of case depth to tooth size
Hardenability and heat treatment
Standard size factors for gear teeth have not yet been established for cases where there
is a detrimental size effect. In such cases AGMA recommends a size factor greater than
unity. If there is no detrimental size effect, use unity.
AGMA has identified and provided a symbol for size factor. Also, AGMA
suggests K s = 1, which makes K s a placeholder in Eqs. (14–15) and (14–16) until
more information is gathered. Following the standard in this manner is a failure to
apply all of your knowledge. From Table 13–1, l = a + b = 2.25/P√. The tooth
thickness t in Fig. 14–6 is given in Sec. 14–1, Eq. (b), as t = 4lx where
x = 3Y/(2P) from Eq. (14–3). From Eq. (6–25) the √
equivalent diameter de of a
rectangular section in bending is de = 0.808 Ft . From Eq. (6–20)
kb = (de /0.3)−0.107 . Noting that K s is the reciprocal of kb , we find the result of all
the algebraic substitution is
√ 0.0535
1
F Y
= 1.192
Ks =
(a)
kb
P
K s can be viewed as Lewis’s geometry incorporated into the Marin size factor in
fatigue. You may set K s = 1, or you may elect to use the preceding Eq. (a). This is a
point to discuss with your instructor. We will use Eq. (a) to remind you that you have a
choice. If K s in Eq. (a) is less than 1, use K s = 1.
14–11
Load-Distribution Factor Km (KH)
The load-distribution factor modified the stress equations to reflect nonuniform distribution of load across the line of contact. The ideal is to locate the gear “midspan” between
two bearings at the zero slope place when the load is applied. However, this is not always
possible. The following procedure is applicable to
• Net face width to pinion pitch diameter ratio F/d ≤ 2
• Gear elements mounted between the bearings
• Face widths up to 40 in
• Contact, when loaded, across the full width of the narrowest member
The load-distribution factor under these conditions is currently given by the face load
distribution factor, Cm f , where
K m = Cm f = 1 + Cmc (C p f C pm + Cma Ce )
(14–30)
738
740
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III. Design of Mechanical
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14. Spur and Helical Gears
Mechanical Engineering Design
where
!
Cmc =
Cp f
1
0.8
for uncrowned teeth
for crowned teeth
⎧
F
⎪
⎪
⎪
− 0.025
⎪
⎪
10d
⎪
⎪
⎨ F
=
− 0.0375 + 0.0125F
⎪
10d
⎪
⎪
⎪
⎪
F
⎪
⎪
− 0.1109 + 0.0207F − 0.000 228F 2
⎩
10d
(14–31)
F ≤ 1 in
1 < F ≤ 17 in
(14–32)
17 < F ≤ 40 in
Note that for values of F/(10d) < 0.05, F/(10d) = 0.05 is used.
!
C pm =
1
1.1
for straddle-mounted pinion with S1 /S < 0.175
for straddle-mounted pinion with S1 /S ≥ 0.175
Cma = A + B F + C F 2
Ce =
⎧
⎨ 0.8
⎩
1
(see Table 14–9 for values of A, B, and C)
for gearing adjusted at assembly, or compatibility
is improved by lapping, or both
for all other conditions
(14–33)
(14–34)
(14–35)
See Fig. 14–10 for definitions of S and S1 for use with Eq. (14–33), and see Fig. 14–11
for graph of Cma .
Table 14–9
Condition
Empirical Constants
A, B, and C for
Eq. (14–34), Face
Width F in Inches∗
Open gearing
Source: ANSI/AGMA
2001-D04.
B
C
0.247
0.0167
−0.765(10−4)
Commercial, enclosed units
0.127
0.0158
−0.930(10−4)
Precision, enclosed units
0.0675
0.0128
−0.926(10−4)
Extraprecision enclosed gear units
0.00360
0.0102
−0.822(10−4)
*See ANSI/AGMA 2101-D04, pp. 20–22, for SI formulation.
Figure 14–10
Definition of distances S and
S1 used in evaluating Cpm,
Eq. (14–33). (ANSI/AGMA
2001-D04.)
A
Centerline of
gear face
Centerline of
bearing
Centerline of
bearing
S
2
S1
S
Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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14. Spur and Helical Gears
Spur and Helical Gears
739
741
0.90
Open gearing
0.80
Mesh alignment factor, Cma
0.70
0.60
Commercial enclosed gear units
Curve 1
0.50
Precision enclosed gear units
0.40
Curve 2
0.30
Curve 3
Extra precision enclosed gear units
0.20
Curve 4
0.10
For determination of Cma , see Eq. (14-34)
0.0
0
5
10
15
20
25
30
35
Face width, F (in)
Figure 14–11
Mesh alignment factor Cma. Curve-fit equations in Table 14–9. (ANSI/AGMA 2001-D04.)
14–12
Hardness-Ratio Factor CH
The pinion generally has a smaller number of teeth than the gear and consequently is subjected to more cycles of contact stress. If both the pinion and the gear are through-hardened,
then a uniform surface strength can be obtained by making the pinion harder than the gear.
A similar effect can be obtained when a surface-hardened pinion is mated with a throughhardened gear. The hardness-ratio factor C H is used only for the gear. Its purpose is to
adjust the surface strengths for this effect. The values of C H are obtained from the equation
C H = 1.0 + A (m G − 1.0)
where
A = 8.98(10−3 )
HB P
HBG
− 8.29(10−3 ) 1.2 ≤
(14–36)
HB P
≤ 1.7
HBG
The terms HB P and HBG are the Brinell hardness (10-mm ball at 3000-kg load) of
the pinion and gear, respectively. The term m G is the speed ratio and is given by
Eq. (14–22). See Fig. 14–12 for a graph of Eq. (14–36). For
HB P
< 1.2,
HBG
A = 0
HB P
> 1.7,
HBG
A = 0.006 98
When surface-hardened pinions with hardnesses of 48 Rockwell C scale (Rockwell
C48) or harder are run with through-hardened gears (180–400 Brinell), a work hardening occurs. The C H factor is a function of pinion surface finish f P and the mating gear
hardness. Figure 14–13 displays the relationships:
C H = 1 + B (450 − HBG )
(14–37)
III. Design of Mechanical
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© The McGraw−Hill
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14. Spur and Helical Gears
Mechanical Engineering Design
Figure 14–12
1.14
1.7
Hardness ratio factor CH
(through-hardened steel).
(ANSI/AGMA 2001-D04.)
Hardness ratio factor, CH
1.12
1.6
1.10
1.5
1.08
1.4
1.3
1.06
HBP
HBG
742
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
Calculated hardness ratio,
740
1.2
1.04
When
HBP
< 1.2,
HBG
Use CH = 1
1.02
1.00
0
2
4
6
8
10
12
14
16
18
20
Single reduction gear ratio mG
Figure 14–13
1.14
Hardness ratio factor, CH
Hardness ratio factor CH
(surface-hardened steel
pinion). (ANSI/AGMA 2001D04.)
Surface Finish of Pinion, fp ,
microinches, Ra
1.16
fp = 16
1.12
fp = 32
1.10
1.08
fp = 64
1.06
1.04
When fp > 64
use CH = 1.0
1.02
1.00
180
200
250
300
350
400
Brinell hardness of the gear, HBG
where B = 0.000 75 exp[−0.0112 f P ] and f P is the surface finish of the pinion
expressed as root-mean-square roughness Ra in μ in.
14–13
Stress Cycle Factors YN and ZN
The AGMA strengths as given in Figs. 14–2 through 14–4, in Tables 14–3 and 14–4 for
bending fatigue, and in Fig. 14–5 and Tables 14–5 and 14–6 for contact-stress fatigue
are based on 107 load cycles applied. The purpose of the load cycle factors Y N and Z N
is to modify the gear strength for lives other than 107 cycles. Values for these factors
are given in Figs. 14–14 and 14–15. Note that for 107 cycles Y N = Z N = 1 on each
graph. Note also that the equations for Y N and Z N change on either side of 107 cycles.
For life goals slightly higher than 107 cycles, the mating gear may be experiencing
fewer than 107 cycles and the equations for (Y N ) P and (Y N )G can be different. The
same comment applies to (Z N ) P and (Z N )G .
Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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741
© The McGraw−Hill
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14. Spur and Helical Gears
Spur and Helical Gears
5.0
Repeatedly applied bending
strength stress-cycle factor YN.
(ANSI/AGMA 2001-D04.)
4.0
Stress cycle factor, YN
Figure 14–14
NOTE: The choice of YN in the shaded
area is influenced by:
YN = 9.4518 N − 0.148
400 HB
3.0 Case carb.
250 HB
Nitrided
2.0
160 HB
YN = 6.1514 N − 0.1192
YN = 4.9404 N − 0.1045
Pitchline velocity
Gear material cleanliness
Residual stress
Material ductility and fracture toughness
YN = 3.517 N − 0.0817
YN = 1.3558 N − 0.0178
YN = 2.3194 N − 0.0538
1.0
0.9
0.8
0.7
0.6
0.5
10 2
743
1.0
0.9
0.8
0.7
0.6
YN = 1.6831 N − 0.0323
10 3
10 4
10 5
10 6
10 7
10 8
10 9
0.5
10 10
Number of load cycles, N
Figure 14–15
5.0
Pitting resistance stress-cycle
factor ZN. (ANSI/AGMA
2001-D04.)
4.0
NOTE: The choice of Z N in the shaded
zone is influenced by:
Stress cycle factor, Z N
3.0
2.0
ZN = 2.466 N − 0.056
Lubrication regime
Failure criteria
Smoothness of operation required
Pitchline velocity
Gear material cleanliness
Material ductility and fracture toughness
Residual stress
ZN = 1.4488 N − 0.023
1.1
1.0
0.9
0.8
0.7
0.6
0.5
102
Nitrided
ZN = 1.249 N − 0.0138
103
104
105
106
107
108
109
1010
Number of load cycles, N
14–14
Reliability Factor KR (YZ)
The reliability factor accounts for the effect of the statistical distributions of material
fatigue failures. Load variation is not addressed here. The gear strengths St and Sc are
based on a reliability of 99 percent. Table 14–10 is based on data developed by the U.S.
Navy for bending and contact-stress fatigue failures.
The functional relationship between K R and reliability is highly nonlinear. When
interpolation is required, linear interpolation is too crude. A log transformation to each
quantity produces a linear string. A least-squares regression fit is
!
KR =
0.658 − 0.0759 ln(1 − R)
0.50 − 0.109 ln(1 − R)
0.5 < R < 0.99
0.99 ≤ R ≤ 0.9999
(14–38)
For cardinal values of R, take K R from the table. Otherwise use the logarithmic interpolation afforded by Eqs. (14–38).
742
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III. Design of Mechanical
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14. Spur and Helical Gears
Mechanical Engineering Design
Table 14–10
Reliability
KR (YZ)
Reliability Factors KR (YZ )
0.9999
1.50
Source: ANSI/AGMA
2001-D04.
0.999
1.25
0.99
1.00
0.90
0.85
0.50
0.70
14–15
Temperature Factor KT (Yθ)
For oil or gear-blank temperatures up to 250°F (120°C), use K T = Yθ = 1.0. For higher
temperatures, the factor should be greater than unity. Heat exchangers may be used to
ensure that operating temperatures are considerably below this value, as is desirable for
the lubricant.
14–16
Rim-Thickness Factor KB
When the rim thickness is not sufficient to provide full support for the tooth root, the
location of bending fatigue failure may be through the gear rim rather than at the tooth
fillet. In such cases, the use of a stress-modifying factor K B or (t R ) is recommended.
This factor, the rim-thickness factor K B , adjusts the estimated bending stress for the
thin-rimmed gear. It is a function of the backup ratio m B ,
tR
ht
mB =
(14–39)
where tR = rim thickness below the tooth, in, and ht = the tooth height. The geometry
is depicted in Fig. 14–16. The rim-thickness factor K B is given by
KB =
Figure 14–16
2.2
Rim thickness factor, KB
Rim thickness factor KB.
(ANSI/AGMA 2001-D04.)
2.4
2.0
For mB < 1.2
KB = 1.6 ln
⎧
⎨
⎩
1.6 ln
2.242
mB
m B < 1.2
m B ≥ 1.2
1
( (
2.242
mB
(14–40)
ht
1.8
1.6
tR
For mB ≥ 1.2
KB = 1.0
1.4
1.2
mB =
tR
ht
1.0
0
0.5 0.6
0.8
1.0
1.2
2
3
Backup ratio, mB
4
5
6
7 8 9 10
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14. Spur and Helical Gears
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Spur and Helical Gears
743
745
Figure 14–16 also gives the value of K B graphically. The rim-thickness factor K B is
applied in addition to the 0.70 reverse-loading factor when applicable.
14–17
Safety Factors SF and SH
The ANSI/AGMA standards 2001-D04 and 2101-D04 contain a safety factor S F
guarding against bending fatigue failure and safety factor S H guarding against pitting
failure.
The definition of S F , from Eq. (14–17), is
SF =
fully corrected bending strength
St Y N /(K T K R )
=
σ
bending stress
(14–41)
where σ is estimated from Eq. (14–15). It is a strength-over-stress definition in a case
where the stress is linear with the transmitted load.
The definition of S H , from Eq. (14–18), is
SH =
Sc Z N C H /(K T K R )
fully corrected contact strength
=
σc
contact stress
(14–42)
when σc is estimated from Eq. (14–16). This, too, is a strength-over-stress definition but
in a case where the stress is not linear with the transmitted load W t .
While the definition of S H does not interfere with its intended function, a caution
is required when comparing S F with S H in an analysis in order to ascertain the nature
and severity of the threat to loss of function. To render S H linear with the transmitted
load, W t it could have been defined as
fully corrected contact strength 2
SH =
(14–43)
contact stress imposed
with the exponent 2 for linear or helical contact, or an exponent of 3 for crowned
teeth (spherical contact). With the definition, Eq. (14–42), compare S F with S H2
(or S H3 for crowned teeth) when trying to identify the threat to loss of function with
confidence.
The role of the overload factor K o is to include predictable excursions of load beyond
W t based on experience. A safety factor is intended to account for unquantifiable elements in addition to K o . When designing a gear mesh, the quantity S F becomes a design
factor (S F )d within the meanings used in this book. The quantity S F evaluated as part of
a design assessment is a factor of safety. This applies equally well to the quantity S H .
14–18
Analysis
Description of the procedure based on the AGMA standard is highly detailed. The best
review is a “road map” for bending fatigue and contact-stress fatigue. Figure 14–17
identifies the bending stress equation, the endurance strength in bending equation, and
the factor of safety S F . Figure 14–18 displays the contact-stress equation, the contact
fatigue endurance strength equation, and the factor of safety S H . When analyzing a gear
problem, this figure is a useful reference.
The following example of a gear mesh analysis is intended to make all the details
presented concerning the AGMA method more familiar.
744
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III. Design of Mechanical
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14. Spur and Helical Gears
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SPUR GEAR BENDING
BASED ON ANSIⲐAGMA 2001-D04
dP =
NP
Pd
V = πdn
12
1 [or Eq. (a), Sec. 14 –10]; p. 739
W t = 33 000 Η
V
Gear
bending
stress
equation
Eq. (14 –15)
␴ = W tKoKvKs
Eq. (14 –30); p. 739
Pd Km KB
J
F
Eq. (14 – 40); p. 744
Fig. 14 – 6; p. 733
Eq. (14 –27); p. 736
Table below
0.99 (St )107
Gear
bending
endurance
strength
equation
Eq. (14–17)
Bending
factor of
safety
Eq. (14–41)
Tables 14 –3, 14 – 4; pp. 728, 729
␴all =
St YN
SF KT KR
Fig. 14 –14; p. 743
Table 14 –10, Eq. (14 –38); pp. 744, 743
1 if T < 250°F
SF =
St YN ⁄ (KT KR)
␴
Remember to compare SF with S 2H when deciding whether bending
or wear is the threat to function. For crowned gears compare SF with S 3H .
Table of Overload Factors, Ko
Driven Machine
Power source
Uniform
Light shock
Medium shock
Uniform Moderate shock Heavy shock
1.00
1.25
1.50
1.25
1.50
1.75
1.75
2.00
2.25
Figure 14–17
Roadmap of gear bending equations based on AGMA standards. (ANSI/AGMA 2001-D04.)
© The McGraw−Hill
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Budynas−Nisbett: Shigley’s
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III. Design of Mechanical
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14. Spur and Helical Gears
Spur and Helical Gears
SPUR GEAR WEAR
BASED ON ANSIⲐAGMA 2001-D04
dP =
NP
Pd
V = πdn
12
1 [or Eq. (a), Sec. 14 –10]; p. 739
Eq. (14 –30); p. 739
1
W t = 33 000 Η
V
Gear
contact
stress
equation
Eq. (14 –16)
(
␴c = Cp W tKoKvKs
K m Cf
dP F I
1⁄2
)
Eq. (14 –23); p. 735
Eq. (14 –27); p. 736
Eq. (14 –13), Table 14 – 8; pp. 724, 737
Table below
0.99 (Sc )107
Gear
contact
endurance
strength
Eq. (14–18)
Tables, 14 –6, 14 –7; pp. 731, 732
Fig. 14 –15; p. 743
␴c,all =
Sc Z N CH
SH KT KR
Section 14 –12, gear only; pp. 741, 742
Table 14 –10, Eqs. (14 –38); pp. 744, 743
1 if T < 250° F
Gear only
Wear
factor of
safety
Eq. (14–42)
SH =
Sc Z N CH ⁄ (KT KR)
␴c
Remember to compare SF with S 2H when deciding whether bending
or wear is the threat to function. For crowned gears compare SF with S 3H .
Table of Overload Factors, Ko
Driven Machine
Power source
Uniform
Light shock
Medium shock
Uniform Moderate shock Heavy shock
1.00
1.25
1.50
1.25
1.50
1.75
1.75
2.00
2.25
Figure 14–18
Roadmap of gear wear equations based on AGMA standards. (ANSI/AGMA 2001-D04.)
745
747
746
748
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14. Spur and Helical Gears
Mechanical Engineering Design
EXAMPLE 14–4
A 17-tooth 20° pressure angle spur pinion rotates at 1800 rev/min and transmits 4 hp to
a 52-tooth disk gear. The diametral pitch is 10 teeth/in, the face width 1.5 in, and the
quality standard is No. 6. The gears are straddle-mounted with bearings immediately
adjacent. The pinion is a grade 1 steel with a hardness of 240 Brinell tooth surface and
through-hardened core. The gear is steel, through-hardened also, grade 1 material, with
a Brinell hardness of 200, tooth surface and core. Poisson’s ratio is 0.30, J P = 0.30,
JG = 0.40, and Young’s modulus is 30(106 ) psi. The loading is smooth because of
motor and load. Assume a pinion life of 108 cycles and a reliability of 0.90, and use
Y N = 1.3558N −0.0178 , Z N = 1.4488N −0.023 . The tooth profile is uncrowned. This is a
commercial enclosed gear unit.
(a) Find the factor of safety of the gears in bending.
(b) Find the factor of safety of the gears in wear.
(c) By examining the factors of safety, identify the threat to each gear and to the mesh.
Solution
There will be many terms to obtain so use Figs. 14–17 and 14–18 as guides to what is
needed.
d P = N P /Pd = 17/10 = 1.7 in
V =
Wt =
dG = 52/10 = 5.2 in
π(1.7)1800
πd P n P
=
= 801.1 ft/min
12
12
33 000(4)
33 000 H
=
= 164.8 lbf
V
801.1
Assuming uniform loading, K o = 1. To evaluate K v , from Eq. (14–28) with a quality
number Q v = 6,
B = 0.25(12 − 6)2/3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
Then from Eq. (14–27) the dynamic factor is
0.8255
√
59.77 + 801.1
= 1.377
Kv =
59.77
To determine the size factor, K s , the Lewis form factor is needed. From Table 14–2,
with N P = 17 teeth, Y P = 0.303. Interpolation for the gear with NG = 52 teeth yields
YG = 0.412. Thus from Eq. (a) of Sec. 14–10, with F = 1.5 in,
0.0535
√
1.5 0.303
= 1.043
(K s ) P = 1.192
10
0.0535
√
1.5 0.412
= 1.052
(K s )G = 1.192
10
The load distribution factor Km is determined from Eq. (14–30), where five terms are
needed. They are, where F = 1.5 in when needed:
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Uncrowned, Eq. (14–30): Cmc = 1,
Eq. (14–32): C p f = 1.5/[10(1.7)] − 0.0375 + 0.0125(1.5) = 0.0695
Bearings immediately adjacent, Eq. (14–33): C pm = 1
Commercial enclosed gear units (Fig. 14–11): Cma = 0.15
Eq. (14–35): Ce = 1
Thus,
K m = 1 + Cmc (C p f C pm + Cma Ce ) = 1 + (1)[0.0695(1) + 0.15(1)] = 1.22
Assuming constant thickness gears, the rim-thickness factor K B = 1. The speed ratio is
m G = NG /N P = 52/17 = 3.059. The load cycle factors given in the problem statement, with N(pinion) = 108 cycles and N(gear) = 108 /m G = 108 /3.059 cycles, are
(Y N ) P = 1.3558(108 )−0.0178 = 0.977
(Y N )G = 1.3558(108 /3.059)−0.0178 = 0.996
From Table 14.10, with a reliability of 0.9, K R = 0.85. From Fig. 14–18, the temperature and surface condition factors are K T = 1 and C f = 1. From Eq. (14–23), with
m N = 1 for spur gears,
I =
cos 20◦ sin 20◦ 3.059
= 0.121
2
3.059 + 1
√
From Table 14–8, C p = 2300 psi.
Next, we need the terms for the gear endurance strength equations. From Table 14–3,
for grade 1 steel with HB P = 240 and HBG = 200, we use Fig. 14–2, which gives
(St ) P = 77.3(240) + 12 800 = 31 350 psi
(St )G = 77.3(200) + 12 800 = 28 260 psi
Similarly, from Table 14–6, we use Fig. 14–5, which gives
(Sc ) P = 322(240) + 29 100 = 106 400 psi
(Sc )G = 322(200) + 29 100 = 93 500 psi
From Fig. 14–15,
(Z N ) P = 1.4488(108 )−0.023 = 0.948
(Z N )G = 1.4488(108 /3.059)−0.023 = 0.973
For the hardness ratio factor CH, the hardness ratio is HB P /HBG = 240/200 = 1.2.
Then, from Sec. 14–12,
A = 8.98(10−3 )(HB P /HBG ) − 8.29(10−3 )
= 8.98(10−3 )(1.2) − 8.29(10−3 ) = 0.002 49
Thus, from Eq. (14–36),
C H = 1 + 0.002 49(3.059 − 1) = 1.005
748
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(a) Pinion tooth bending. Substituting the appropriate terms for the pinion into
Eq. (14–15) gives
10 1.22 (1)
Pd K m K B
= 164.8(1)1.377(1.043)
(σ ) P = W t K o K v K s
F
J
1.5
0.30
P
= 6417 psi
Answer
Substituting the appropriate terms for the pinion into Eq. (14–41) gives
St Y N /(K T K R )
31 350(0.977)/[1(0.85)]
= 5.62
=
(S F ) P =
σ
6417
P
Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15)
gives
(σ )G = 164.8(1)1.377(1.052)
10 1.22(1)
= 4854 psi
1.5 0.40
Substituting the appropriate terms for the gear into Eq. (14–41) gives
Answer
(S F )G =
28 260(0.996)/[1(0.85)]
= 6.82
4854
(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16)
gives
K m C f 1/2
(σc ) P = C p W t K o K v K s
dP F I P
1/2
1
1.22
= 2300 164.8(1)1.377(1.043)
= 70 360 psi
1.7(1.5) 0.121
Answer
Substituting the appropriate terms for the pinion into Eq. (14–42) gives
Sc Z N /(K T K R )
106 400(0.948)/[1(0.85)]
= 1.69
=
(S H ) P =
σc
70 360
P
Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is Ks. Thus,
(K s )G 1/2
1.052 1/2
(σc ) P =
70 360 = 70 660 psi
(σc )G =
(K s ) P
1.043
Substituting the appropriate terms for the gear into Eq. (14–42) with C H = 1.005 gives
Answer
(S H )G =
93 500(0.973)1.005/[1(0.85)]
= 1.52
70 660
(c) For the pinion, we compare (SF)P with (S H )2P , or 5.73 with 1.692 = 2.86, so the
threat in the pinion is from wear. For the gear, we compare (SF)G with (S H )2G , or 6.96
with 1.522 = 2.31, so the threat in the gear is also from wear.
There are perspectives to be gained from Ex. 14–4. First, the pinion is overly strong
in bending compared to wear. The performance in wear can be improved by surfacehardening techniques, such as flame or induction hardening, nitriding, or carburizing
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751
and case hardening, as well as shot peening. This in turn permits the gearset to be made
smaller. Second, in bending, the gear is stronger than the pinion, indicating that both the
gear core hardness and tooth size could be reduced; that is, we may increase P and
reduce diameter of the gears, or perhaps allow a cheaper material. Third, in wear,
surface strength equations have the ratio (Z N )/K R . The values of (Z N ) P and (Z N )G are
affected by gear ratio m G . The designer can control strength by specifying surface
hardness. This point will be elaborated later.
Having followed a spur-gear analysis in detail in Ex. 14–4, it is timely to analyze
a helical gearset under similar circumstances to observe similarities and differences.
EXAMPLE 14–5
A 17-tooth 20◦ normal pitch-angle helical pinion with a right-hand helix angle of 30◦
rotates at 1800 rev/min when transmitting 4 hp to a 52-tooth helical gear. The normal
diametral pitch is 10 teeth/in, the face width is 1.5 in, and the set has a quality number
of 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion
and gear are made from a through-hardened steel with surface and core hardnesses of
240 Brinell on the pinion and surface and core hardnesses of 200 Brinell on the gear.
The transmission is smooth, connecting an electric motor and a centrifugal pump.
Assume a pinion life of 108 cycles and a reliability of 0.9 and use the upper curves in
Figs. 14–14 and 14–15.
(a) Find the factors of safety of the gears in bending.
(b) Find the factors of safety of the gears in wear.
(c) By examining the factors of safety identify the threat to each gear and to the mesh.
Solution
All of the parameters in this example are the same as in Ex. 14–4 with the exception that
we are using helical gears. Thus, several terms will be the same as Ex. 14–4. The reader
should verify that the following terms remain unchanged: K o = 1, Y P = 0.303, YG =
0.412, m G = 3.059, (K s ) P = 1.043, (K s )√G = 1.052, (Y N ) P = 0.977, (Y N )G = 0.996,
K R = 0.85, K T = 1, C f = 1, C p = 2300 psi, (St ) P = 31 350 psi, (St )G = 28 260 psi,
(Sc ) P = 106 380 psi, (Sc )G = 93 500 psi, (Z N ) P = 0.948, (Z N )G = 0.973, and C H =
1.005.
For helical gears, the transverse diametral pitch, given by Eq. (13–18), is
Pt = Pn cos ψ = 10 cos 30◦ = 8.660 teeth/in
Thus, the pitch diameters are d P = N P /Pt = 17/8.660 = 1.963 in and dG = 52/
8.660 = 6.005 in. The pitch-line velocity and transmitted force are
π(1.963)1800
πd P n P
=
= 925 ft/min
12
12
33 000(4)
33 000H
=
= 142.7 lbf
Wt =
V
925
V =
As in Ex. 14–4, for the dynamic factor, B = 0.8255 and A = 59.77. Thus, Eq. (14–27)
gives
0.8255
√
59.77 + 925
= 1.404
Kv =
59.77
The geometry factor I for helical gears requires a little work. First, the transverse pressure
750
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14. Spur and Helical Gears
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angle is given by Eq. (13–19)
tan φn
tan 20o
= tan−1
= 22.80o
φt = tan−1
cos ψ
cos 30o
The radii of the pinion and gear are r P = 1.963/2 = 0.9815 in and r G = 6.004/2 =
3.002 in, respectively. The addendum is a = 1/Pn = 1/10 = 0.1, and the base-circle
radii of the pinion and gear are given by Eq. (13–6) with φ = φt :
(rb ) P = r P cos φt = 0.9815 cos 22.80◦ = 0.9048 in
(rb )G = 3.002 cos 22.80◦ = 2.767 in
From Eq. (14–25), the surface strength geometry factor
Z=
(0.9815 + 0.1)2 − 0.90482 +
(3.004 + 0.1)2 − 2.7692
− (0.9815 + 3.004) sin 22.80◦
= 0.5924 + 1.4027 − 1.544 4 = 0.4507 in
Since the first two terms are less than 1.544 4, the equation for Z stands. From
Eq. (14–24) the normal circular pitch p N is
p N = pn cos φn =
π
π
cos 20◦ = 0.2952 in
cos 20◦ =
Pn
10
From Eq. (14–21), the load sharing ratio
mN =
pN
0.2952
=
= 0.6895
0.95Z
0.95(0.4507)
Substituting in Eq. (14–23), the geometry factor I is
I =
sin 22.80◦ cos 22.80◦ 3.06
= 0.195
2(0.6895)
3.06 + 1
From Fig. 14–7, geometry factors J P = 0.45 and JG = 0.54. Also from Fig. 14–8 the
J-factor multipliers are 0.94 and 0.98, correcting J P and JG to
J P = 0.45(0.94) = 0.423
JG = 0.54(0.98) = 0.529
The load-distribution factor K m is estimated from Eq. (14–32):
Cp f =
1.5
− 0.0375 + 0.0125(1.5) = 0.0577
10(1.963)
with Cmc = 1, C pm = 1, Cma = 0.15 from Fig. 14–11, and Ce = 1. Therefore, from
Eq. (14–30),
K m = 1 + (1)[0.0577(1) + 0.15(1)] = 1.208
(a) Pinion tooth bending. Substituting the appropriate terms into Eq. (14–15) using Pt
gives
8.66 1.208(1)
Pt K m K B
= 142.7(1)1.404(1.043)
(σ ) P = W t K o K v K s
F
J
1.5 0.423
P
= 3445 psi
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14. Spur and Helical Gears
Spur and Helical Gears
751
753
Substituting the appropriate terms for the pinion into Eq. (14–41) gives
Answer
(S F ) P =
St Y N /(K T K R )
σ
=
P
31 350(0.977)/[1(0.85)]
= 10.5
3445
Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives
(σ )G = 142.7(1)1.404(1.052)
8.66 1.208(1)
= 2779 psi
1.5 0.529
Substituting the appropriate terms for the gear into Eq. (14–41) gives
Answer
(S F )G =
28 260(0.996)/[1(0.85)]
= 11.9
2779
(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into
Eq. (14–16) gives
K m C f 1/2
(σc ) P = C p W t K o K v K s
dP F I P
1/2
1
1.208
= 2300 142.7(1)1.404(1.043)
= 48 230 psi
1.963(1.5) 0.195
Answer
Substituting the appropriate terms for the pinion into Eq. (14–42) gives
Sc Z N /(K T K R )
106 400(0.948)/[1(0.85)]
= 2.46
=
(S H ) P =
σc
48 230
P
Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is Ks. Thus,
(K s )G 1/2
1.052 1/2
(σc ) P =
48 230 = 48 440 psi
(σc )G =
(K s ) P
1.043
Substituting the appropriate terms for the gear into Eq. (14–42) with C H = 1.005 gives
Answer
(S H )G =
93 500(0.973)1.005/[1(0.85)]
= 2.22
48 440
(c) For the pinion we compare S F with S H2 , or 10.5 with 2.462 = 6.05, so the threat in
the pinion is from wear. For the gear we compare S F with S H2 , or 11.9 with 2.222 = 4.93,
so the threat is also from wear in the gear. For the meshing gearset wear controls.
It is worthwhile to compare Ex. 14–4 with Ex. 14–5. The spur and helical gearsets
were placed in nearly identical circumstances. The helical gear teeth are of greater
length because of the helix and identical face widths. The pitch diameters of the helical
gears are larger. The J factors and the I factor are larger, thereby reducing stresses. The
result is larger factors of safety. In the design phase the gearsets in Ex. 14–4 and
Ex. 14–5 can be made smaller with control of materials and relative hardnesses.
Now that examples have given the AGMA parameters substance, it is time to examine some desirable (and necessary) relationships between material properties of spur
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14. Spur and Helical Gears
Mechanical Engineering Design
gears in mesh. In bending, the AGMA equations are displayed side by side:
Pd K m K B
Pd K m K B
σG = W t K o K v K s
σP = W t Ko Kv Ks
F
J
F
J
P
G
St Y N /(K T K R )
St Y N /(K T K R )
(S F ) P =
(S F )G =
σ
σ
P
G
Equating the factors of safety, substituting for stress and strength, canceling identical
terms (K s virtually equal or exactly equal), and solving for (St )G gives
(St )G = (St ) P
(Y N ) P J P
(Y N )G JG
(a)
The stress-cycle factor Y N comes from Fig. 14–14, where for a particular hardness,
β
Y N = α N β . For the pinion, (Y N ) P = α N P , and for the gear, (Y N )G = α(N P /m G )β .
Substituting these into Eq. (a) and simplifying gives
β
(St )G = (St ) P m G
JP
JG
(14–44)
Normally, m G > 1 and JG > J P , so equation (14–44) shows that the gear can be less
strong (lower Brinell hardness) than the pinion for the same safety factor.
EXAMPLE 14–6
Solution
In a set of spur gears, a 300-Brinell 18-tooth 16-pitch 20◦ full-depth pinion meshes with
a 64-tooth gear. Both gear and pinion are of grade 1 through-hardened steel. Using β =
−0.023, what hardness can the gear have for the same factor of safety?
For through-hardened grade 1 steel the pinion strength (St ) P is given in Fig. 14–2:
(St ) P = 77.3(300) + 12 800 = 35 990 psi
From Fig. 14–6 the form factors are J P = 0.32 and JG = 0.41. Equation (14–44) gives
−0.023
64
0.32
= 27 280 psi
(St )G = 35 990
18
0.41
Use the equation in Fig. 14–2 again.
(HB )G =
Answer
27 280 − 12 800
= 187 Brinell
77.3
The AGMA contact-stress equations also are displayed side by side:
K m C f 1/2
K m C f 1/2
(σc )G = C p W t K o K v K s
(σc ) P = C p W t K o K v K s
dP F I P
dP F I G
(S H ) P =
Sc Z N /(K T K R )
σc
(S H )G =
P
Sc Z N C H /(K T K R )
σc
G
Equating the factors of safety, substituting the stress relations, and canceling identical
Budynas−Nisbett: Shigley’s
Mechanical Engineering
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III. Design of Mechanical
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© The McGraw−Hill
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14. Spur and Helical Gears
Spur and Helical Gears
753
755
terms including K s gives, after solving for (Sc )G ,
(Z N ) P
1
1
β
= (SC ) P m G
(Sc )G = (Sc ) P
(Z N )G C H G
CH G
β
where, as in the development of Eq. (14–44), (Z N ) P /(Z N )G = m G and the value of β
for wear comes from Fig. 14–15. Since C H is so close to unity, it is usually neglected;
therefore
β
(Sc )G = (Sc ) P m G
EXAMPLE 14–7
Solution
(14–45)
For β = −0.056 for a through-hardened steel, grade 1, continue Ex. 14–6 for wear.
From Fig. 14–5,
(Sc ) P = 322(300) + 29 100 = 125 700 psi
From Eq. (14–45),
(Sc )G = (Sc ) P
64
18
−0.056
(HB )G =
Answer
= 125 700
64
18
−0.056
= 117 100 psi
117 100 − 29 200
= 273 Brinell
322
which is slightly less than the pinion hardness of 300 Brinell.
Equations (14–44) and (14–45) apply as well to helical gears.
14–19
Design of a Gear Mesh
A useful decision set for spur and helical gears includes
• Function: load, speed, reliability, life, K o
• Unquantifiable risk: design factor n d
• Tooth system: φ, ψ, addendum, dedendum, root fillet radius
• Gear ratio m G , N p , NG
•
•
•
•
Quality number Q v
Diametral pitch Pd
Face width F
Pinion material, core hardness, case hardness
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎭
a priori decisions
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎭
design decisions
• Gear material, core hardness, case hardness
The first item to notice is the dimensionality of the decision set. There are four design
decision categories, eight different decisions if you count them separately. This is a larger
number than we have encountered before. It is important to use a design strategy that is
convenient in either longhand execution or computer implementation. The design decisions
have been placed in order of importance (impact on the amount of work to be redone in
iterations). The steps are, after the a priori decisions have been made,
754
756
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
Mechanical Engineering Design
• Choose a diametral pitch.
• Examine implications on face width, pitch diameters, and material properties. If not
satisfactory, return to pitch decision for change.
• Choose a pinion material and examine core and case hardness requirements. If not
satisfactory, return to pitch decision and iterate until no decisions are changed.
• Choose a gear material and examine core and case hardness requirements. If not
satisfactory, return to pitch decision and iterate until no decisions are changed.
With these plan steps in mind, we can consider them in more detail.
First select a trial diametral pitch.
Pinion bending:
• Select a median face width for this pitch, 4π/P
• Find the range of necessary ultimate strengths
•
•
•
•
Choose a material and a core hardness
Find face width to meet factor of safety in bending
Choose face width
Check factor of safety in bending
Gear bending:
• Find necessary companion core hardness
• Choose a material and core hardness
• Check factor of safety in bending
Pinion wear:
• Find necessary Sc and attendant case hardness
• Choose a case hardness
• Check factor of safety in wear
Gear wear:
• Find companion case hardness
• Choose a case hardness
• Check factor of safety in wear
Completing this set of steps will yield a satisfactory design. Additional designs
with diametral pitches adjacent to the first satisfactory design will produce several
among which to choose. A figure of merit is necessary in order to choose the best.
Unfortunately, a figure of merit in gear design is complex in an academic environment
because material and processing cost vary. The possibility of using a process depends
on the manufacturing facility if gears are made in house.
After examining Ex. 14–4 and Ex. 14–5 and seeing the wide range of factors of
safety, one might entertain the notion of setting all factors of safety equal.9 In steel
9
In designing gears it makes sense to define the factor of safety in wear as (S)2H for uncrowned teeth, so that
there is no mix-up. ANSI, in the preface to ANSI/AGMA 2001-D04 and 2101-D04, states “the use is completely voluntary. . . does not preclude anyone from using . . . procedures . . . not conforming to the standards.”
Budynas−Nisbett: Shigley’s
Mechanical Engineering
Design, Eighth Edition
III. Design of Mechanical
Elements
14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
755
Spur and Helical Gears
757
gears, wear is usually controlling and (S H ) P and (S H )G can be brought close to equality. The use of softer cores can bring down (S F ) P and (S F )G , but there is value in keeping them higher. A tooth broken by bending fatigue not only can destroy the gear set,
but can bend shafts, damage bearings, and produce inertial stresses up- and downstream
in the power train, causing damage elsewhere if the gear box locks.
EXAMPLE 14–8
Solution
Design a 4:1 spur-gear reduction for a 100-hp, three-phase squirrel-cage induction
motor running at 1120 rev/min. The load is smooth, providing a reliability of 0.95 at 109
revolutions of the pinion. Gearing space is meager. Use Nitralloy 135M, grade 1 material to keep the gear size small. The gears are heat-treated first then nitrided.
Make the a priori decisions:
•
•
•
•
•
•
Function: 100 hp, 1120 rev/min, R = 0.95, N = 109 cycles, K o = 1
Design factor for unquantifiable exingencies: n d = 2
Tooth system: φn = 20◦
Tooth count: N P = 18 teeth, NG = 72 teeth (no interference)
Quality number: Q v = 6, use grade 1 material
Assume m B ≥ 1.2 in Eq. (14–40), K B = 1
Pitch: Select a trial diametral pitch of Pd = 4 teeth/in. Thus, d P = 18/4 = 4.5 in and
dG = 72/4 = 18 in. From Table 14–2, Y P = 0.309, YG = 0.4324 (interpolated). From
Fig. 14–6, J P = 0.32, JG = 0.415.
V =
π(4.5)1120
πd P n P
=
= 1319 ft/min
12
12
Wt =
33 000(100)
33 000H
=
= 2502 lbf
V
1319
From Eqs. (14–28) and (14–27),
B = 0.25(12 − Q v )2/3 = 0.25(12 − 6)2/3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
Kv =
0.8255
√
59.77 + 1319
= 1.480
59.77
From Eq. (14–38), K R = 0.658 − 0.0759 ln(1 − 0.95) = 0.885. From Fig. 14–14,
(Y N ) P = 1.3558(109 )−0.0178 = 0.938
(Y N )G = 1.3558(109 /4)−0.0178 = 0.961
From Fig. 14–15,
(Z N ) P = 1.4488(109 )−0.023 = 0.900
(Z N )G = 1.4488(109 /4)−0.023 = 0.929
756
758
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III. Design of Mechanical
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14. Spur and Helical Gears
Mechanical Engineering Design
From the recommendation after Eq. (14–8), 3 p ≤ F ≤ 5 p. Try F = 4 p = 4π/P =
4π/4 = 3.14 in. From Eq. (a), Sec. 14–10,
0.0535
√ 0.0535
√
F Y
3.14 0.309
= 1.192
= 1.140
K s = 1.192
P
4
From Eqs. (14–31), (14–33), (14–35), Cmc = C pm = Ce = 1. From Fig. 14–11, Cma =
0.175 for commercial enclosed gear units. From Eq. (14–32), F/(10d P ) = 3.14/
[10(4.5)] = 0.0698. Thus,
C p f = 0.0698 − 0.0375 + 0.0125(3.14) = 0.0715
From Eq. (14–30),
K m = 1 + (1)[0.0715(1) + 0.175(1)] = 1.247
√
From Table 14–8, for steel gears, C p = 2300 psi. From Eq. (14–23), with m G = 4 and
m N = 1,
I =
cos 20o sin 20o 4
= 0.1286
2
4+1
Pinion tooth bending. With the above estimates of K s and K m from the trial diametral
pitch, we check to see if the mesh width F is controlled by bending or wear considerations. Equating Eqs. (14–15) and (14–17), substituting n d W t for W t , and solving for
the face width (F)bend necessary to resist bending fatigue, we obtain
(F)bend = n d W t K o K v K s Pd
Km K B KT K R
JP
St Y N
(1)
Equating Eqs. (14–16) and (14–18), substituting n d W t for W t , and solving for the face
width (F)wear necessary to resist wear fatigue, we obtain
(F)wear =
Cp Z N
Sc K T K R
2
nd W t Ko Kv Ks
Km C f
dP I
(2)
From Table 14–5 the hardness range of Nitralloy 135M is Rockwell C32–36 (302–335
Brinell). Choosing a midrange hardness as attainable, using 320 Brinell. From
Fig. 14–4,
St = 86.2(320) + 12 730 = 40 310 psi
Inserting the numerical value of St in Eq. (1) to estimate the face width gives
(F)bend = 2(2502)(1)1.48(1.14)4
1.247(1)(1)0.885
= 3.08 in
0.32(40 310)0.938
From Table 14–6 for Nitralloy 135M, Sc = 170 000 psi. Inserting this in Eq. (2), we
find
(F)wear =
2300(0.900)
170 000(1)0.885
2
2(2502)1(1.48)1.14
1.247(1)
= 3.44 in
4.5(0.1286)
Budynas−Nisbett: Shigley’s
Mechanical Engineering
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III. Design of Mechanical
Elements
© The McGraw−Hill
Companies, 2008
14. Spur and Helical Gears
Spur and Helical Gears
Decision
757
759
Make face width 3.50 in. Correct K s and K m :
0.0535
√
3.50 0.309
= 1.147
K s = 1.192
4
3.50
F
=
= 0.0778
10d P
10(4.5)
C p f = 0.0778 − 0.0375 + 0.0125(3.50) = 0.0841
K m = 1 + (1)[0.0841(1) + 0.175(1)] = 1.259
The bending stress induced by W t in bending, from Eq. (14–15), is
(σ ) P = 2502(1)1.48(1.147)
4 1.259(1)
= 19 100 psi
3.50 0.32
The AGMA factor of safety in bending of the pinion, from Eq. (14–41), is
(S F ) P =
Decision
40 310(0.938)/[1(0.885)]
= 2.24
19 100
Gear tooth bending. Use cast gear blank because of the 18-in pitch diameter. Use the
same material, heat treatment, and nitriding. The load-induced bending stress is in the
ratio of J P /JG . Then
(σ )G = 19 100
0.32
= 14 730 psi
0.415
The factor of safety of the gear in bending is
(S F )G =
40 310(0.961)/[1(0.885)]
= 2.97
14 730
Pinion tooth wear. The contact stress, given by Eq. (14–16), is
1/2
1
1.259
= 118 000 psi
(σc ) P = 2300 2502(1)1.48(1.147)
4.5(3.5) 0.129
The factor of safety from Eq. (14–42), is
(S H ) P =
170 000(0.900)/[1(0.885)]
= 1.465
118 000
By our definition of factor of safety, pinion bending is (S F ) P = 2.24, and wear is
(S H )2P = (1.465)2 = 2.15.
Gear tooth wear. The hardness of the gear and pinion are the same. Thus, from
Fig. 14–12, C H = 1, the contact stress on the gear is the same as the pinion, (σc )G =
118 000 psi. The wear strength is also the same, Sc = 170 000 psi. The factor of safety
of the gear in wear is
(S H )G =
170 000(0.929)/[1(0.885)]
= 1.51
118 000
So, for the gear in bending, (S F )G = 2.97, and wear (S H )2G = (1.51)2 = 2.29.
758
760
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III. Design of Mechanical
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14. Spur and Helical Gears
© The McGraw−Hill
Companies, 2008
Mechanical Engineering Design
Rim. Keep m B ≥ 1.2. The whole depth is h t = addendum + dedendum = 1/Pd +
1.25/Pd = 2.25/Pd = 2.25/4 = 0.5625 in. The rim thickness t R is
t R ≥ m B h t = 1.2(0.5625) = 0.675 in
In the design of the gear blank, be sure the rim thickness exceeds 0.675 in; if it does
not, review and modify this mesh design.
This design example showed a satisfactory design for a four-pitch spur-gear mesh.
Material could be changed, as could pitch. There are a number of other satisfactory
designs, thus a figure of merit is needed to identify the best.
One can appreciate that gear design was one of the early applications of the digital
computer to mechanical engineering. A design program should be interactive, presenting results of calculations, pausing for a decision by the designer, and showing the consequences of the decision, with a loop back to change a decision for the better. The
program can be structured in totem-pole fashion, with the most influential decision at
the top, then tumbling down, decision after decision, ending with the ability to change
the current decision or to begin again. Such a program would make a fine class project.
Troubleshooting the coding will reinforce your knowledge, adding flexibility as well as
bells and whistles in subsequent terms.
Standard gears may not be the most economical design that meets the functional
requirements, because no application is standard in all respects.10 Methods of designing custom gears are well-understood and frequently used in mobile equipment to provide good weight-to-performance index. The required calculations including optimizations are within the capability of a personal computer.
PROBLEMS
Because gearing problems can be difficult, the problems are presented by section.
Section 14–1
14–1
A steel spur pinion has a pitch of 6 teeth/in, 22 full-depth teeth, and a 20◦ pressure angle. The
pinion runs at a speed of 1200 rev/min and transmits 15 hp to a 60-tooth gear. If the face width
is 2 in, estimate the bending stress.
14–2
A steel spur pinion has a diametral pitch of 12 teeth/in, 16 teeth cut full-depth with a 20◦ pressure
angle, and a face width of 34 in. This pinion is expected to transmit 1.5 hp at a speed of 700 rev/min.
Determine the bending stress.
14–3
A steel spur pinion has a module of 1.25 mm, 18 teeth cut on the 20◦ full-depth system, and a
face width of 12 mm. At a speed of 1800 rev/min, this pinion is expected to carry a steady load
of 0.5 kW. Determine the resulting bending stress.
14–4
A steel spur pinion has 15 teeth cut on the 20◦ full-depth system with a module of 5 mm and a
face width of 60 mm. The pinion rotates at 200 rev/min and transmits 5 kW to the mating steel
gear. What is the resulting bending stress?
10
See H. W. Van Gerpen, C. K. Reece, and J. K. Jensen, Computer Aided Design of Custom Gears,
Van Gerpen–Reece Engineering, Cedar Falls, Iowa, 1996.